### The Mizar article:

### On the Decomposition of Finite Sequences

**by****Andrzej Trybulec**

- Received May 24, 1995
Copyright (c) 1995 Association of Mizar Users

- MML identifier: FINSEQ_6
- [ MML identifier index ]

environ vocabulary FINSEQ_1, REALSET1, FINSEQ_5, RELAT_1, BOOLE, FUNCT_1, FINSEQ_4, ARYTM_1, RLSUB_2, RFINSEQ, FINSEQ_6, ORDINAL2, ARYTM; notation TARSKI, XBOOLE_0, ENUMSET1, SUBSET_1, ORDINAL1, ORDINAL2, NUMBERS, XCMPLX_0, XREAL_0, NAT_1, RELAT_1, RELSET_1, FUNCT_1, FINSEQ_1, FINSEQ_4, RFINSEQ, FINSEQ_5, REALSET1, TOPREAL1; constructors REAL_1, NAT_1, MATRIX_2, FINSEQ_5, ENUMSET1, RFINSEQ, TOPREAL1, REALSET1, FINSEQ_4, INT_1, MEMBERED, XBOOLE_0; clusters RELSET_1, FINSEQ_5, FUNCT_1, GOBOARD4, FINSEQ_1, INT_1, REALSET1, ZFMISC_1, XBOOLE_0, ORDINAL2, ARYTM_3; requirements REAL, NUMERALS, BOOLE, SUBSET, ARITHM; definitions TARSKI, XBOOLE_0; theorems TOPREAL1, CQC_THE1, NAT_1, REAL_1, AXIOMS, FINSEQ_1, FINSEQ_4, FINSEQ_3, RLVECT_1, FINSEQ_2, FUNCT_1, ZFMISC_1, FINSEQ_5, RELAT_1, REAL_2, RFINSEQ, TARSKI, ENUMSET1, PARTFUN2, SPPOL_1, GRFUNC_1, HAHNBAN, FSM_1, REALSET1, GROUP_5, SQUARE_1, INT_1, XBOOLE_0, XBOOLE_1, XCMPLX_1; begin :: Preliminaries reserve x,y,z for set; definition let x,y,z; cluster <*x,y,z*> -> non trivial; coherence proof len <*x,y,z*> = 3 by FINSEQ_1:62; hence <*x,y,z*> is non trivial by SPPOL_1:2; end; end; definition let f be non empty FinSequence; cluster Rev f -> non empty; coherence proof dom f <> {} by RELAT_1:64; hence Rev f is non empty by FINSEQ_5:60,RELAT_1:60; end; end; Lm1: rng <*x,y*> = { x,y } proof <*x,y*> = <*x*>^<*y*> by FINSEQ_1:def 9; hence rng<*x,y*> = rng<*x*> \/ rng<*y*> by FINSEQ_1:44 .= {x} \/ rng<*y*> by FINSEQ_1:55 .= {x} \/ {y} by FINSEQ_1:55 .= {x,y} by ENUMSET1:41; end; Lm2: rng <*x,y,z*> = { x,y,z } proof <*x,y,z*> = <*x,y*>^<*z*> by FINSEQ_1:60; hence rng<*x,y,z*> = rng<*x,y*> \/ rng<*z*> by FINSEQ_1:44 .= {x,y} \/ rng<*z*> by Lm1 .= {x,y} \/ {z} by FINSEQ_1:55 .= {x,y,z} by ENUMSET1:43; end; begin reserve f,f1,f2,f3 for FinSequence, p,p1,p2,p3 for set, i,k for Nat; canceled 2; theorem Th3: for X being set, i st X c= Seg i & 1 in X holds (Sgm X).1 = 1 proof let X be set, i such that A1: X c= Seg i and A2: 1 in X; A3: rng Sgm X = X by A1,FINSEQ_1:def 13; then consider x such that A4: x in dom Sgm X and A5: (Sgm X).x = 1 by A2,FUNCT_1:def 5; reconsider j = x as Nat by A4; A6: j >= 1 by A4,FINSEQ_3:27; A7: rng Sgm X c= NAT by FINSEQ_1:def 4; assume A8: (Sgm X).1 <> 1; Sgm X <> {} by A1,A2,FINSEQ_1:72; then len Sgm X <> 0 by FINSEQ_1:25; then len Sgm X >= 1 by RLVECT_1:99; then 1 in dom Sgm X by FINSEQ_3:27; then A9: (Sgm X).1 in rng Sgm X & (Sgm X).j in rng Sgm X by A4,FUNCT_1: def 5; then reconsider k1 = (Sgm X).1 as Nat by A7; A10: k1 >= 1 by A1,A3,A9,FINSEQ_1:3; A11: j <= len Sgm X by A4,FINSEQ_3:27; 1 < j by A5,A6,A8,REAL_1:def 5; hence contradiction by A1,A5,A10,A11, FINSEQ_1:def 13; end; theorem Th4: for f being FinSequence holds k in dom f & (for i st 1 <= i & i < k holds f.i <> f.k) implies (f.k)..f = k proof let f be FinSequence; assume that A1: k in dom f and A2: for i st 1 <= i & i < k holds f.i <> f.k; A3: f.k in rng f by A1,FUNCT_1:def 5; then A4: f.(f.k..f) = f.k by FINSEQ_4:29; A5: f.k..f <= k by A1,FINSEQ_4:34; assume f.k..f <> k; then 1 <= f.k.. f & f.k..f < k by A3,A5,FINSEQ_4:31,REAL_1:def 5; hence contradiction by A2,A4; end; theorem Th5: <*p1,p2*>| Seg 1 = <*p1*> proof len<*p1,p2*> = 2 by FINSEQ_1:61; then 1 in dom<*p1,p2*> by FINSEQ_3:27; then A1: Seg 1 c= dom<*p1,p2*> by FINSEQ_1:4,ZFMISC_1:37; set f = <*p1,p2*>| Seg 1; A2: dom f = dom<*p1,p2*> /\ Seg 1 by RELAT_1:90 .= Seg 1 by A1,XBOOLE_1:28; then reconsider f as FinSequence by FINSEQ_1:def 2; A3: len f = 1 by A2,FINSEQ_1:def 3; 1 in dom f by A2,FINSEQ_1:4,TARSKI:def 1; then f.1 = <*p1,p2*>.1 by FUNCT_1:70 .= p1 by FINSEQ_1:61; hence thesis by A3,FINSEQ_1:57; end; theorem Th6: <*p1,p2,p3*>| Seg 1 = <*p1*> proof len<*p1,p2,p3*> = 3 by FINSEQ_1:62; then 1 in dom<*p1,p2,p3*> by FINSEQ_3:27; then A1: Seg 1 c= dom<*p1,p2,p3*> by FINSEQ_1:4,ZFMISC_1:37; set f = <*p1,p2,p3*>| Seg 1; A2: dom f = dom<*p1,p2,p3*> /\ Seg 1 by RELAT_1:90 .= Seg 1 by A1,XBOOLE_1:28 ; then reconsider f as FinSequence by FINSEQ_1:def 2; A3: len f = 1 by A2,FINSEQ_1:def 3; 1 in dom f by A2,FINSEQ_1:4,TARSKI:def 1; then f.1 = <*p1,p2,p3*>.1 by FUNCT_1:70 .= p1 by FINSEQ_1:62; hence thesis by A3,FINSEQ_1:57; end; theorem Th7: <*p1,p2,p3*>| Seg 2 = <*p1,p2*> proof len<*p1,p2,p3*> = 3 by FINSEQ_1:62; then 1 in dom<*p1,p2,p3*> & 2 in dom<*p1,p2,p3*> by FINSEQ_3:27; then A1: Seg 2 c= dom<*p1,p2,p3*> by FINSEQ_1:4,ZFMISC_1:38; set f = <*p1,p2,p3*>| Seg 2; A2: dom f = dom<*p1,p2,p3*> /\ Seg 2 by RELAT_1:90 .= Seg 2 by A1,XBOOLE_1:28 ; then reconsider f as FinSequence by FINSEQ_1:def 2; A3: len f = 2 by A2,FINSEQ_1:def 3; then 1 in dom f by FINSEQ_3:27; then A4: f.1 = <*p1,p2,p3*>.1 by FUNCT_1:70 .= p1 by FINSEQ_1:62; 2 in dom f by A3,FINSEQ_3:27; then f.2 = <*p1,p2,p3*>.2 by FUNCT_1:70 .= p2 by FINSEQ_1:62; hence thesis by A3,A4,FINSEQ_1:61; end; theorem Th8: p in rng f1 implies p..(f1^f2) = p..f1 proof assume A1: p in rng f1; then A2: f1.(p..f1) = p by FINSEQ_4:29; A3: p..f1 in dom f1 by A1,FINSEQ_4:30; then A4: (f1^f2).(p..f1) = p by A2,FINSEQ_1:def 7; A5: dom f1 c= dom(f1^f2) by FINSEQ_1:39; now let i such that A6: 1 <= i and A7: i < p..f1; p..f1 <= len f1 by A1,FINSEQ_4:31; then i <= len f1 by A7,AXIOMS:22; then A8: i in dom f1 by A6,FINSEQ_3:27; A9: (f1^f2).(p..f1) = f1.(p..f1) by A3,FINSEQ_1:def 7; (f1^f2).i = f1.i by A8,FINSEQ_1:def 7; hence (f1^f2).i <> (f1^f2).(p..f1) by A2,A7,A8,A9,FINSEQ_4:34; end; hence p..(f1^f2) = p..f1 by A3,A4,A5,Th4; end; theorem Th9: p in rng f2 \ rng f1 implies p..(f1^f2) = len f1 + p..f2 proof assume A1: p in rng f2 \ rng f1; then A2: p in rng f2 by XBOOLE_0:def 4; then A3: f2.(p..f2) = p by FINSEQ_4:29; A4: p..f2 in dom f2 by A2,FINSEQ_4:30; then A5: (f1^f2).(len f1 + p..f2) = p by A3,FINSEQ_1:def 7; A6: len f1 + p..f2 in dom(f1^f2) by A4,FINSEQ_1:41; now let i such that A7: 1 <= i and A8: i < len f1 + p..f2; per cases; suppose i <= len f1; then A9: i in dom f1 by A7,FINSEQ_3:27; assume (f1^f2).i = (f1^f2).(len f1 + p..f2); then f1.i = p by A5,A9,FINSEQ_1:def 7; then p in rng f1 by A9,FUNCT_1:def 5; hence contradiction by A1,XBOOLE_0:def 4; suppose A10: i > len f1; then reconsider j = i - len f1 as Nat by INT_1:18; A11: i = j + len f1 by XCMPLX_1:27; j > 0 by A10,SQUARE_1:11; then A12: 1 <= j by RLVECT_1:99; A13: j < p..f2 by A8,A11,AXIOMS:24; j <= p..f2 & p..f2 <= len f2 by A2,A8,A11,AXIOMS:24,FINSEQ_4:31; then j <= len f2 by AXIOMS:22; then A14: j in dom f2 by A12,FINSEQ_3:27; A15: (f1^f2).(len f1 + p..f2) = f2.(p..f2) by A4,FINSEQ_1:def 7; (f1^f2).i = f2.j by A11,A14,FINSEQ_1:def 7; hence (f1^f2).i <> (f1^f2).(len f1 + p..f2) by A3,A13,A14,A15,FINSEQ_4:34 ; end; hence p..(f1^f2) = len f1 + p..f2 by A5,A6,Th4; end; theorem Th10: p in rng f1 implies (f1^f2)|--p = (f1|--p)^f2 proof assume A1: p in rng f1; rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:44; then A2: p in rng(f1^f2) by A1,XBOOLE_0:def 2; A3: p..f1 = p..(f1^f2) by A1,Th8; A4: len(f1|--p) = len f1 - p..f1 by A1,FINSEQ_4:def 7; then A5: len((f1|--p)^f2) = len f1 - p..f1 + len f2 by FINSEQ_1:35 .= len f1 + len f2 - p..f1 by XCMPLX_1:29 .= len(f1^f2) - p..(f1^f2) by A3,FINSEQ_1:35; A6: len f1 = len(f1|--p) + p..f1 by A4,XCMPLX_1:27; now let k; assume A7: k in dom((f1|--p)^f2); per cases by A7,FINSEQ_1:38; suppose A8: k in dom(f1|--p); then 1 <= k & k <= k + p..f1 by FINSEQ_3:27,NAT_1:29; then A9: 1 <= k + p..f1 by AXIOMS:22; A10: k <= len(f1|--p) by A8,FINSEQ_3:27; len(f1|--p) = len f1 - p..f1 by A1,FINSEQ_4:def 7; then len(f1|--p) + p..f1 = len f1 by XCMPLX_1:27; then k + p..f1 <= len f1 by A10,AXIOMS:24; then A11: k + p..f1 in dom f1 by A9,FINSEQ_3:27; thus ((f1|--p)^f2).k = (f1|--p).k by A8,FINSEQ_1:def 7 .= f1.(k + p..f1) by A1,A8,FINSEQ_4:def 7 .= (f1^f2).(k + p..(f1^f2)) by A3,A11,FINSEQ_1:def 7; suppose ex n being Nat st n in dom f2 & k = len(f1|--p) + n; then consider n being Nat such that A12: n in dom f2 and A13: k = len(f1|--p) + n; thus ((f1|--p)^f2).k = f2.n by A12,A13,FINSEQ_1:def 7 .= (f1^f2).(len f1 + n) by A12,FINSEQ_1:def 7 .= (f1^f2).(k + p..(f1^f2)) by A3,A6,A13,XCMPLX_1:1; end; hence (f1^f2)|--p = (f1|--p)^f2 by A2,A5,FINSEQ_4:def 7; end; theorem Th11: p in rng f2 \ rng f1 implies (f1^f2)|--p = f2|--p proof assume A1: p in rng f2 \ rng f1; then A2: p in rng f2 by XBOOLE_0:def 4; rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:44; then A3: p in rng(f1^f2) by A2,XBOOLE_0:def 2; A4: len f1 + p..f2 = p..(f1^f2) by A1,Th9; A5: len(f2|--p) = len f2 - p..f2 by A2,FINSEQ_4:def 7 .= len f1 + len f2 - (len f1 + p..f2) by XCMPLX_1:32 .= len(f1^f2) - p..(f1^f2) by A4,FINSEQ_1:35; now let k; assume A6: k in dom(f2|--p); then 1 <= k & k <= k + p..f2 by FINSEQ_3:27,NAT_1:29; then A7: 1 <= k + p..f2 by AXIOMS:22; A8: k <= len(f2|--p) by A6,FINSEQ_3:27; len(f2|--p) = len f2 - p..f2 by A2,FINSEQ_4:def 7; then len(f2|--p) + p..f2 = len f2 by XCMPLX_1:27; then k + p..f2 <= len f2 by A8,AXIOMS:24; then A9: k + p..f2 in dom f2 by A7,FINSEQ_3:27; thus (f2|--p).k = f2.(k + p..f2) by A2,A6,FINSEQ_4:def 7 .= (f1^f2).(len f1 + (k + p..f2)) by A9,FINSEQ_1:def 7 .= (f1^f2).(k + p..(f1^f2)) by A4,XCMPLX_1:1; end; hence (f1^f2)|--p = f2|--p by A3,A5,FINSEQ_4:def 7; end; theorem Th12: f1 c= f1^f2 proof A1: dom f1 c= dom(f1^f2) by FINSEQ_1:39; for x st x in dom f1 holds f1.x = (f1^f2).x by FINSEQ_1:def 7; hence f1 c= f1^f2 by A1,GRFUNC_1:8; end; theorem Th13: for A being set st A c= dom f1 holds (f1^f2)|A = f1 | A proof let A be set; f1 c= f1^f2 by Th12; hence thesis by HAHNBAN:2; end; theorem Th14: p in rng f1 implies (f1^f2)-|p = f1-|p proof assume A1: p in rng f1; then consider n being Nat such that A2: n = p..f1 - 1 and A3: f1 | Seg n = f1 -| p by FINSEQ_4:def 6; rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:44; then A4: p in rng(f1^f2) by A1,XBOOLE_0:def 2; A5: n = p..(f1^f2) - 1 by A1,A2,Th8; n + 1 = p..f1 by A2,XCMPLX_1:27; then A6: n <= p..f1 by NAT_1:29; p..f1 <= len f1 by A1,FINSEQ_4:31; then n <= len f1 by A6,AXIOMS:22; then Seg n c= Seg len f1 by FINSEQ_1:7; then Seg n c= dom f1 by FINSEQ_1:def 3; then (f1^f2) | Seg n = f1 | Seg n by Th13; hence (f1^f2)-|p = f1-|p by A3,A4,A5,FINSEQ_4:def 6; end; definition let f1; let i be natural number; cluster f1|Seg i -> FinSequence-like; coherence by FINSEQ_1:19; end; theorem Th15: f1 c= f2 implies f3^f1 c= f3^f2 proof assume A1: f1 c= f2; A2: dom(f3^f1) c= dom(f3^f2) proof let x; assume A3: x in dom(f3^f1); then reconsider i = x as Nat; per cases by A3,FINSEQ_1:38; suppose A4: i in dom f3; dom f3 c= dom(f3^f2) by FINSEQ_1:39; hence x in dom(f3^f2) by A4; suppose ex n being Nat st n in dom f1 & i=len f3 + n; then consider k such that A5: k in dom f1 and A6: i = len f3 + k; dom f1 c= dom f2 by A1,RELAT_1:25; hence x in dom(f3^f2) by A5,A6,FINSEQ_1:41; end; for x st x in dom(f3^f1) holds (f3^f1).x = (f3^f2).x proof let x; assume A7: x in dom(f3^f1); then reconsider i = x as Nat; per cases by A7,FINSEQ_1:38; suppose A8: i in dom f3; hence (f3^f1).x = f3.i by FINSEQ_1:def 7 .= (f3^f2).x by A8,FINSEQ_1:def 7; suppose ex n being Nat st n in dom f1 & i=len f3 + n; then consider k such that A9: k in dom f1 and A10: i = len f3 + k; A11: dom f1 c= dom f2 by A1,RELAT_1:25; thus (f3^f1).x = f1.k by A9,A10,FINSEQ_1:def 7 .= f2.k by A1,A9,GRFUNC_1:8 .= (f3^f2).x by A9,A10,A11,FINSEQ_1:def 7; end; hence f3^f1 c= f3^f2 by A2,GRFUNC_1:8; end; theorem Th16: (f1^f2)|Seg(len f1 + i) = f1^(f2|Seg i) proof A1: dom((f1^f2)|Seg(len f1 + i)) = dom(f1^f2) /\ Seg(len f1 + i) by RELAT_1:90; f2|Seg i c= f2 by RELAT_1:88; then f1^(f2|Seg i) c= f1^f2 by Th15; then A2: dom(f1^(f2|Seg i)) c= dom(f1^f2) by RELAT_1:25; dom(f1^(f2|Seg i)) c= Seg(len f1 + i) proof let x; assume A3: x in dom(f1^(f2|Seg i)); then reconsider j = x as Nat; per cases by A3,FINSEQ_1:38; suppose j in dom f1; then A4: j in Seg len f1 by FINSEQ_1:def 3; len f1 <= len f1 + i by NAT_1:29; then Seg len f1 c= Seg(len f1 + i) by FINSEQ_1:7; hence x in Seg(len f1 + i) by A4; suppose ex n being Nat st n in dom(f2|Seg i) & j=len f1 + n; then consider k such that A5: k in dom(f2|Seg i) and A6: j = len f1 + k; dom(f2|Seg i) = dom f2 /\ Seg i by RELAT_1:90; then k in Seg i by A5,XBOOLE_0:def 3; then k <= i by FINSEQ_1:3; then A7: j <= len f1 + i by A6,AXIOMS:24; 1 <= k & k <= j by A5,A6,FINSEQ_3:27,NAT_1:29; then 1 <= j by AXIOMS:22; hence x in Seg(len f1 + i) by A7,FINSEQ_1:3; end; then A8: dom(f1^(f2|Seg i)) c= dom(f1^f2) /\ Seg(len f1 + i) by A2,XBOOLE_1:19; dom(f1^f2) /\ Seg(len f1 + i) c= dom(f1^(f2|Seg i)) proof let x; assume A9: x in dom(f1^f2) /\ Seg(len f1 + i); then A10: x in dom(f1^f2) by XBOOLE_0:def 3; reconsider j = x as Nat by A9; per cases by A10,FINSEQ_1:38; suppose A11: j in dom f1; dom f1 c= dom(f1^(f2|Seg i)) by FINSEQ_1:39; hence x in dom(f1^(f2|Seg i)) by A11; suppose ex n being Nat st n in dom f2 & j=len f1 + n; then consider k such that A12: k in dom f2 and A13: j = len f1 + k; j in Seg(len f1 + i) by A9,XBOOLE_0:def 3; then j <= len f1 + i by FINSEQ_1:3; then A14: k <= i by A13,REAL_1:53; 1 <= k by A12,FINSEQ_3:27; then A15: k in Seg i by A14,FINSEQ_1:3; dom(f2|Seg i) = dom f2 /\ Seg i by RELAT_1:90; then k in dom(f2|Seg i) by A12,A15,XBOOLE_0:def 3; hence x in dom(f1^(f2|Seg i)) by A13,FINSEQ_1:41; end; then A16: dom((f1^f2)|Seg(len f1 + i)) = dom(f1^(f2|Seg i)) by A1,A8, XBOOLE_0:def 10 ; now let k; assume A17: k in dom((f1^f2)|Seg(len f1 + i)); then A18: 1 <= k by FINSEQ_3:27; per cases; suppose k <= len f1; then A19: k in dom f1 by A18,FINSEQ_3:27; thus ((f1^f2)|Seg(len f1 + i)).k = (f1^f2).k by A17,FUNCT_1:70 .= f1.k by A19,FINSEQ_1:def 7 .= (f1^(f2|Seg i)).k by A19,FINSEQ_1:def 7; suppose A20: k > len f1; then reconsider j = k - len f1 as Nat by INT_1:18; A21: k = len f1 + j by XCMPLX_1:27; j > 0 by A20,SQUARE_1:11; then A22: 1 <= j by RLVECT_1:99; k in Seg(len f1 + i) by A1,A17,XBOOLE_0:def 3; then k <= len f1 + i by FINSEQ_1:3; then j <= i by A21,REAL_1:53; then A23: j in Seg i by A22,FINSEQ_1:3; A24: not k in dom f1 by A20,FINSEQ_3:27; k in dom(f1^f2) by A1,A17,XBOOLE_0:def 3; then consider n being Nat such that A25: n in dom f2 and A26: k =len f1 + n by A24,FINSEQ_1:38; A27: j in dom f2 by A21,A25,A26,XCMPLX_1:2; dom(f2|Seg i) = (dom f2) /\ Seg i by RELAT_1:90; then A28: j in dom(f2|Seg i) by A23,A27,XBOOLE_0:def 3; thus ((f1^f2)|Seg(len f1 + i)).k = (f1^f2).k by A17,FUNCT_1:70 .= f2.j by A21,A27,FINSEQ_1:def 7 .= (f2|Seg i).j by A28,FUNCT_1:70 .= (f1^(f2|Seg i)).k by A21,A28,FINSEQ_1:def 7; end; hence (f1^f2)|Seg(len f1 + i) = f1^(f2|Seg i) by A16,FINSEQ_1:17; end; theorem Th17: p in rng f2 \ rng f1 implies (f1^f2)-|p = f1^(f2-|p) proof assume A1: p in rng f2 \ rng f1; then A2: p in rng f2 by XBOOLE_0:def 4; then consider n being Nat such that A3: n = p..f2 - 1 and A4: f2 | Seg n = f2 -| p by FINSEQ_4:def 6; rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:44; then A5: p in rng(f1^f2) by A2,XBOOLE_0:def 2; p..(f1^f2) = len f1 + p..f2 by A1,Th9; then A6: len f1 + n = p..(f1^f2) - 1 by A3,XCMPLX_1:29; (f1^f2) | Seg(len f1 + n) = f1^(f2-|p) by A4,Th16; hence (f1^f2)-|p = f1^(f2-|p) by A5,A6,FINSEQ_4:def 6; end; canceled; theorem Th19: f1^f2 just_once_values p implies p in rng f1 \+\ rng f2 proof assume A1: f1^f2 just_once_values p; then A2: p in rng(f1^f2) by FINSEQ_4:7; A3: rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:44; now assume A4: p in rng f1 /\ rng f2; then p in rng f1 by XBOOLE_0:def 3; then (f1^f2)|--p = (f1|--p)^f2 by Th10; then A5: not p in rng((f1|--p)^f2) by A1,FINSEQ_4:60; rng((f1|--p)^f2) = rng(f1|--p) \/ rng f2 by FINSEQ_1:44; then not p in rng f2 by A5,XBOOLE_0:def 2; hence contradiction by A4,XBOOLE_0:def 3; end; then p in (rng f1 \/ rng f2) \ rng f1 /\ rng f2 by A2,A3,XBOOLE_0:def 4; hence p in rng f1 \+\ rng f2 by XBOOLE_1:101; end; theorem f1^f2 just_once_values p & p in rng f1 implies f1 just_once_values p proof assume that A1: f1^f2 just_once_values p and A2: p in rng f1; A3: not p in rng((f1^f2)|--p) by A1,FINSEQ_4:60; (f1^f2)|--p = (f1|--p)^f2 by A2,Th10; then rng((f1^f2)|--p) = rng(f1|--p) \/ rng f2 by FINSEQ_1:44; then not p in rng(f1|--p) by A3,XBOOLE_0:def 2; hence f1 just_once_values p by A2,FINSEQ_4:60; end; theorem Th21: rng f is non trivial implies f is non trivial proof assume that A1: rng f is non trivial and A2: f is trivial; consider y being set such that A3: f = {y} by A1,A2,REALSET1:def 12,RELAT_1:60; y in f by A3,TARSKI:def 1; then consider y1,y2 being set such that A4: y = [y1,y2] by RELAT_1:def 1; rng f = {y2} by A3,A4,RELAT_1:23; hence contradiction by A1,REALSET1:def 12; end; theorem Th22: p..<*p*> = 1 proof dom<*p*> = Seg 1 by FINSEQ_1:55; then A1: 1 in dom<*p*> by FINSEQ_1:4,TARSKI:def 1; A2: <*p*>.1 = p by FINSEQ_1:57; for i st 1<=i & i<1 holds <*p*>.i <> <*p*>.1; hence thesis by A1,A2,Th4; end; theorem Th23: p1..<*p1,p2*> = 1 proof len<*p1,p2*> = 2 by FINSEQ_1:61; then A1: 1 in dom<*p1,p2*> by FINSEQ_3:27; A2: <*p1,p2*>.1 = p1 by FINSEQ_1:61; for i st 1<=i & i<1 holds <*p1,p2*>.i <> <*p1,p2*>.1; hence thesis by A1,A2,Th4; end; theorem Th24: p1 <> p2 implies p2..<*p1,p2*> = 2 proof assume A1: p1 <> p2; 2 <= len<*p1,p2*> by FINSEQ_1:61; then A2: 2 in dom<*p1,p2*> by FINSEQ_3:27; A3: <*p1,p2*>.2 = p2 by FINSEQ_1:61; A4: <*p1,p2*>.1 = p1 by FINSEQ_1:61; now let i; assume A5: 1<=i; assume i<1+1; then i <= 1 by NAT_1:38; hence <*p1,p2*>.i <> <*p1,p2*>.2 by A1,A3,A4,A5,AXIOMS:21; end; hence thesis by A2,A3,Th4; end; theorem Th25: p1..<*p1,p2,p3*> = 1 proof len<*p1,p2,p3*> = 3 by FINSEQ_1:62; then A1: 1 in dom<*p1,p2,p3*> by FINSEQ_3:27; A2: <*p1,p2,p3*>.1 = p1 by FINSEQ_1:62; for i st 1<=i & i<1 holds <*p1,p2,p3*>.i <> <*p1,p2,p3*>.1; hence thesis by A1,A2,Th4; end; theorem Th26: p1 <> p2 implies p2..<*p1,p2,p3*> = 2 proof assume A1: p1 <> p2; len<*p1,p2,p3*> = 3 by FINSEQ_1:62; then A2: 2 in dom<*p1,p2,p3*> by FINSEQ_3:27; A3: <*p1,p2,p3*>.2 = p2 by FINSEQ_1:62; A4: <*p1,p2,p3*>.1 = p1 by FINSEQ_1:62; now let i; assume A5: 1<=i; assume i<1+1; then i <= 1 by NAT_1:38; hence <*p1,p2,p3*>.i <> <*p1,p2,p3*>.2 by A1,A3,A4,A5,AXIOMS:21; end; hence thesis by A2,A3,Th4; end; theorem Th27: p1 <> p3 & p2 <> p3 implies p3..<*p1,p2,p3*> = 3 proof assume A1: p1 <> p3 & p2 <> p3; 3 <= len<*p1,p2,p3*> by FINSEQ_1:62; then A2: 3 in dom<*p1,p2,p3*> by FINSEQ_3:27; A3: <*p1,p2,p3*>.3 = p3 by FINSEQ_1:62; A4: <*p1,p2,p3*>.2 = p2 & <*p1,p2,p3*>.1 = p1 by FINSEQ_1:62; now let i; assume A5: 1<=i; assume i<2+1; then i <> 0 & i <= 2 by A5,NAT_1:38; hence <*p1,p2,p3*>.i <> <*p1,p2,p3*>.3 by A1,A3,A4,CQC_THE1:3; end; hence thesis by A2,A3,Th4; end; theorem Th28: for f being FinSequence holds Rev(<*p*>^f) = Rev f ^ <*p*> proof let f be FinSequence; thus Rev(<*p*>^f) = Rev f ^ Rev<*p*> by FINSEQ_5:67 .= Rev f ^ <*p*> by FINSEQ_5:63; end; theorem Th29: for f being FinSequence holds Rev Rev f = f proof let f be FinSequence; A1: len f = len Rev f by FINSEQ_5:def 3; now let i; assume A2: i in dom f; then i in dom Rev f by FINSEQ_5:60; then i <= len Rev f by FINSEQ_3:27; then reconsider j = len(Rev f) - i as Nat by INT_1:18; j + 1 + i = 1 + (j + i) by XCMPLX_1:1 .= len(Rev f) + 1 by XCMPLX_1:27; then A3: j + 1 in dom Rev f by A1,A2,FINSEQ_5:62; thus f.i = f.(len f -(len(Rev f) - i)) by A1,XCMPLX_1:18 .= f.(len f -(len(Rev f) - i) - 1 + 1) by XCMPLX_1:27 .= f.(len f -(len(Rev f) - i + 1) + 1) by XCMPLX_1:36 .= (Rev f).(len(Rev f) - i + 1) by A3,FINSEQ_5:def 3; end; hence Rev Rev f = f by A1,FINSEQ_5:def 3; end; theorem Th30: x <> y implies <*x,y*> -| y = <*x*> proof assume A1: x <> y; rng<*x,y*> = { x,y } by Lm1; then A2: y in rng<*x,y*> by TARSKI:def 2; y..<*x,y*> = 1+1 by A1,Th24; then 1 = y..<*x,y*>-1; hence <*x,y*> -| y = <*x,y*>| Seg 1 by A2,FINSEQ_4:45 .= <*x*> by Th5; end; theorem Th31: x <> y implies <*x,y,z*> -| y = <*x*> proof assume A1: x <> y; rng<*x,y,z*> = { x,y,z } by Lm2; then A2: y in rng<*x,y,z*> by ENUMSET1:14; y..<*x,y,z*> = 1+1 by A1,Th26; then 1 = y..<*x,y,z*>-1; hence <*x,y,z*> -| y = <*x,y,z*>| Seg 1 by A2,FINSEQ_4:45 .= <*x*> by Th6; end; theorem Th32: x <> z & y <> z implies <*x,y,z*> -| z = <*x,y*> proof assume A1: x <> z & y <> z; rng<*x,y,z*> = { x,y,z } by Lm2; then A2: z in rng<*x,y,z*> by ENUMSET1:14; z..<*x,y,z*> = 2+1 by A1,Th27; then 2 = z..<*x,y,z*>-1; hence <*x,y,z*> -| z = <*x,y,z*>| Seg 2 by A2,FINSEQ_4:45 .= <*x,y*> by Th7; end; theorem <*x,y*>|--x = <*y*> proof x in { x,y } by TARSKI:def 2; then A1: x in rng<*x,y*> by Lm1; A2: x..<*x,y*> = 1 by Th23; then len<*y*> + x..<*x,y*> = 1 + 1 by FINSEQ_1:57 .= len<*x,y*> by FINSEQ_1:61; then A3: len<*y*> = len<*x,y*> - x..<*x,y*> by XCMPLX_1:26; now let k; assume k in dom<*y*>; then k in Seg 1 by FINSEQ_1:55; then A4: k = 1 by FINSEQ_1:4,TARSKI:def 1; hence <*y*>.k = y by FINSEQ_1:57 .= <*x,y*>.(k + x..<*x,y*>) by A2,A4,FINSEQ_1:61; end; hence thesis by A1,A3,FINSEQ_4:def 7; end; theorem Th34: x <> y implies <*x,y,z*>|--y = <*z*> proof assume A1: x <> y; y in { x,y,z } by ENUMSET1:14; then A2: y in rng<*x,y,z*> by Lm2; A3: y..<*x,y,z*> = 2 by A1,Th26; then len<*z*> + y..<*x,y,z*> = 1 + 2 by FINSEQ_1:57 .= len<*x,y,z*> by FINSEQ_1:62; then A4: len<*z*> = len<*x,y,z*> - y..<*x,y,z*> by XCMPLX_1:26; now let k; assume k in dom<*z*>; then k in Seg 1 by FINSEQ_1:55; then A5: k = 1 by FINSEQ_1:4,TARSKI:def 1; hence <*z*>.k = z by FINSEQ_1:57 .= <*x,y,z*>.(k + y..<*x,y,z*>) by A3,A5,FINSEQ_1:62; end; hence thesis by A2,A4,FINSEQ_4:def 7; end; theorem <*x,y,z*>|--x = <*y,z*> proof x in { x,y,z } by ENUMSET1:14; then A1: x in rng<*x,y,z*> by Lm2; A2: x..<*x,y,z*> = 1 by Th25; then len<*y,z*> + x..<*x,y,z*> = 2 + 1 by FINSEQ_1:61 .= len<*x,y,z*> by FINSEQ_1:62; then A3: len<*y,z*> = len<*x,y,z*> - x..<*x,y,z*> by XCMPLX_1:26; A4: len<*y,z*> = 2 by FINSEQ_1:61; now let k; assume k in dom<*y,z*>; then A5: k in Seg 2 by A4,FINSEQ_1:def 3; per cases by A5,FINSEQ_1:4,TARSKI:def 2; suppose A6: k = 1; hence <*y,z*>.k = y by FINSEQ_1:61 .= <*x,y,z*>.(k + x..<*x,y,z*>) by A2,A6,FINSEQ_1:62; suppose A7: k = 2; hence <*y,z*>.k = z by FINSEQ_1:61 .= <*x,y,z*>.(k + x..<*x,y,z*>) by A2,A7,FINSEQ_1:62; end; hence thesis by A1,A3,FINSEQ_4:def 7; end; theorem Th36: <*z*>|--z = {} & <*z*>-|z = {} proof z in { z } by TARSKI:def 1; then A1: z in rng<*z*> by FINSEQ_1:56; A2: z..<*z*> = 1 & len<*z*> = 1 by Th22,FINSEQ_1:56; hence <*z*>|--z = {} by A1,FINSEQ_4:64; thus <*z*>-|z = {} by A1,A2,FINSEQ_4:52; end; theorem Th37: x <> y implies <*x,y*> |-- y = {} proof y in { x,y } by TARSKI:def 2; then A1: y in rng<*x,y*> by Lm1; assume x <> y; then y..<*x,y*> = 2 & len<*x,y*> = 2 by Th24,FINSEQ_1:61; hence thesis by A1,FINSEQ_4:64; end; theorem Th38: x <> z & y <> z implies <*x,y,z*> |-- z = {} proof z in { x,y,z } by ENUMSET1:14; then A1: z in rng<*x,y,z*> by Lm2; assume x <> z & y <> z; then z..<*x,y,z*> = 3 & len<*x,y,z*> = 3 by Th27,FINSEQ_1:62; hence thesis by A1,FINSEQ_4:64; end; theorem Th39: x in rng f & y in rng(f-|x) implies f-|x-|y = f-|y proof assume that A1: x in rng f and A2: y in rng(f-|x); thus f-|y = ((f -| x) ^ <* x *> ^ (f |-- x))-|y by A1,FINSEQ_4:66 .= ((f -| x) ^ (<* x *> ^ (f |-- x)))-|y by FINSEQ_1:45 .= f-|x-|y by A2,Th14; end; theorem Th40: not x in rng f1 implies x..(f1^<*x*>^f2) = len f1 + 1 proof assume A1: not x in rng f1; x in {x} by TARSKI:def 1; then A2: x in rng<*x*> by FINSEQ_1:55; then x in rng<*x*> \ rng f1 by A1,XBOOLE_0:def 4; then A3: (f1^<*x*>)|--x = <*x*>|--x by Th11 .= {} by Th36; rng(f1^<*x*>) = rng f1 \/ rng<*x*> by FINSEQ_1:44; then A4: x in rng(f1^<*x*>) by A2,XBOOLE_0:def 2; then A5: len(f1^<*x*>) - x..(f1^<*x*>) = len ((f1^<*x*>)|--x) by FINSEQ_4:def 7 .= 0 by A3,FINSEQ_1:25; thus x..(f1^<*x*>^f2) = x..(f1^<*x*>) by A4,Th8 .= len(f1^<*x*>) by A5,XCMPLX_1:15 .= len f1 + len<*x*> by FINSEQ_1:35 .= len f1 + 1 by FINSEQ_1:56; end; theorem Th41: f just_once_values x implies x..f + x..Rev f = len f + 1 proof assume A1: f just_once_values x; then A2: x in rng f by FINSEQ_4:7; then A3: f = (f -| x) ^ <* x *> ^ (f |-- x) by FINSEQ_4:66; then A4: len f = len((f -| x) ^ <* x *>) + len(f |-- x) by FINSEQ_1:35 .= len(f -| x) + len<* x *> + len(f |-- x) by FINSEQ_1:35 .= len(f -| x) + 1 + len(f |-- x) by FINSEQ_1:56; A5: len(f -| x) + 1 = x..f - 1 + 1 by A2,FINSEQ_4:46 .= x..f by XCMPLX_1:27; not x in rng(f|--x) by A1,FINSEQ_4:60; then A6: not x in rng Rev(f |-- x) by FINSEQ_5:60; Rev f = Rev(f |-- x) ^ Rev((f -| x) ^ <* x *>) by A3,FINSEQ_5:67 .= Rev(f |-- x) ^ (<*x*> ^ Rev(f -| x)) by FINSEQ_5:66 .= Rev(f |-- x) ^ <*x*> ^ Rev(f -| x) by FINSEQ_1:45; then x..Rev f = len Rev(f |-- x) + 1 by A6,Th40; hence x..f + x..Rev f = len(f -| x) + 1 + len Rev(f |-- x) + 1 by A5,XCMPLX_1: 1 .= len f + 1 by A4,FINSEQ_5:def 3; end; theorem Th42: f just_once_values x implies Rev(f-|x) = Rev f |--x proof assume A1: f just_once_values x; then A2: x in rng f by FINSEQ_4:7; then A3: x in rng Rev f by FINSEQ_5:60; A4: x..f + x..Rev f = len f + 1 by A1,Th41; A5: len Rev(f-|x) = len(f-|x) by FINSEQ_5:def 3; A6: len(f-|x) = x..f - 1 by A2,FINSEQ_4:46 .= len f - x..Rev f by A4,XCMPLX_1:33 .= len Rev f - x..Rev f by FINSEQ_5:def 3; now let k; assume A7: k in dom Rev(f-|x); then A8: k in dom(f-|x) by FINSEQ_5:60; consider m being Nat such that m = x..f - 1 and A9: f-|x = f | Seg m by A2,FINSEQ_4:def 6; A10: x..f - 1 - k + 1 = x..f - k - 1 + 1 by XCMPLX_1:21 .= x..f - k by XCMPLX_1:27; A11: len(f-|x) = x..f - 1 by A2,FINSEQ_4:46; then k <= x..f - 1 by A8,FINSEQ_3:27; then A12: k + 1 <= x..f by REAL_1:84; then A13: 1 <= x..f - k by REAL_1:84; then 0 <= x..f - k by AXIOMS:22; then k <= x..f by REAL_2:105; then A14: x..f - k is Nat by INT_1:18; A15: 1 <= k by A7,FINSEQ_3:27; k <= k + x..Rev f by NAT_1:29; then A16: 1 <= k + x..Rev f by A15,AXIOMS:22; x..f - k <= x..f - 1 by A15,REAL_1:92; then A17: x..f - 1 - k + 1 in dom(f | Seg m) by A9,A10,A11,A13,A14, FINSEQ_3:27; k < x..f by A12,NAT_1:38; then k + x..Rev f < len f + 1 by A4,REAL_1:53; then k + x..Rev f <= len f by NAT_1:38; then k + x..Rev f <= len Rev f by FINSEQ_5:def 3; then A18: k + x..Rev f in dom Rev f by A16,FINSEQ_3:27; thus (Rev(f-|x)).k = (f-|x).(x..f - 1 - k + 1) by A7,A11,FINSEQ_5:def 3 .= f.(x..f - 1 - k + 1) by A9,A17,FUNCT_1:70 .= f.(len f - x..Rev f - k + 1) by A4,XCMPLX_1:33 .= f.(len f -(k + x..Rev f) + 1) by XCMPLX_1:36 .= (Rev f).(k + x..Rev f) by A18,FINSEQ_5:def 3; end; hence Rev(f-|x) = Rev f |--x by A3,A5,A6,FINSEQ_4:def 7; end; theorem f just_once_values x implies Rev f just_once_values x proof assume A1: f just_once_values x; then A2: x in rng f by FINSEQ_4:7; then A3: x in rng Rev f by FINSEQ_5:60; not x in rng(f-|x) by A2,FINSEQ_4:49; then not x in rng Rev(f-|x) by FINSEQ_5:60; then not x in rng(Rev f |-- x) by A1,Th42; hence Rev f just_once_values x by A3,FINSEQ_4:60; end; begin reserve D for non empty set, p,p1,p2,p3 for Element of D, f,f1,f2 for FinSequence of D; theorem Th44: p in rng f implies f-:p = (f -| p)^<*p*> proof assume p in rng f; hence (f -| p)^<*p*> = f|(p..f) by FINSEQ_5:27 .= f-:p by FINSEQ_5:def 1; end; theorem Th45: p in rng f implies f:-p = <*p*>^(f |-- p) proof assume p in rng f; hence <*p*>^(f |-- p) = <*p*>^(f/^(p..f)) by FINSEQ_5:38 .= f:-p by FINSEQ_5:def 2; end; theorem Th46: f <> {} implies f/.1 in rng f proof assume f <> {}; then 1 in dom f by FINSEQ_5:6; hence thesis by PARTFUN2:4; end; theorem Th47: f <> {} implies f/.1..f = 1 proof assume A1: f <> {}; f"{f/.1} c= dom f by RELAT_1:167; then A2: f"{f/.1} c= Seg len f by FINSEQ_1:def 3; A3: 1 in dom f by A1,FINSEQ_5:6; f/.1 in {f/.1} by TARSKI:def 1; then A4: 1 in (f qua Relation of NAT,D)"{f/.1} by A3,PARTFUN2:44; thus f/.1..f = Sgm(f"{f/.1}).1 by FINSEQ_4:def 5 .= 1 by A2,A4,Th3; end; theorem Th48: f <> {} & f/.1 = p implies f-:p = <*p*> & f:-p = f proof assume A1: f <> {} & f/.1 = p; then A2: p in rng f by Th46; p..f = 1 by A1,Th47; then A3: f -| p = {} by A2,FINSEQ_4:52; hence f-:p = {}^<*p*> by A2,Th44 .= <*p*> by FINSEQ_1:47; thus f = {} ^ <* p *> ^ (f |-- p) by A2,A3,FINSEQ_4:66 .= <* p *> ^ (f |-- p) by FINSEQ_1:47 .= f:-p by A2,Th45; end; theorem Th49: (<*p1*>^f)/^1 = f proof A1: (<*p1*>^f)/.1 = p1 by FINSEQ_5:16; A2: dom<*p1*> c= dom(<*p1*>^f) by FINSEQ_1:39; 1 in Seg 1 by FINSEQ_1:4,TARSKI:def 1; then 1 in dom<*p1*> by FINSEQ_1:55; then <*(<*p1*>^f)/.(0+1)*>^((<*p1*>^f)/^1) = (<*p1*>^f)/^0 by A2,FINSEQ_5:34 .= <*p1*>^f by FINSEQ_5:31; hence thesis by A1,FINSEQ_1:46; end; theorem Th50: <*p1,p2*>/^1 = <*p2*> proof <*p1,p2*> = <*p1*>^<*p2*> by FINSEQ_1:def 9; hence thesis by Th49; end; theorem Th51: <*p1,p2,p3*>/^1 = <*p2,p3*> proof <*p1,p2,p3*> = <*p1*>^<*p2,p3*> by FINSEQ_1:60; hence thesis by Th49; end; theorem Th52: k in dom f & (for i st 1 <= i & i < k holds f/.i <> f/.k) implies f/.k..f = k proof assume that A1: k in dom f and A2: for i st 1 <= i & i < k holds f/.i <> f/.k; A3: f/.k in rng f by A1,PARTFUN2:4; A4: f/.k..f <= k by A1,FINSEQ_5:42; assume f/.k..f <> k; then 1 <= f/.k.. f & f/.k..f < k by A3,A4,FINSEQ_4:31,REAL_1:def 5; then f/.(f/.k..f) <> f/.k by A2; hence contradiction by A3,FINSEQ_5:41; end; theorem Th53: p1 <> p2 implies <*p1,p2*>-:p2 = <*p1,p2*> proof assume A1: p1 <> p2; rng<*p1,p2*> = { p1,p2 } by Lm1; then p2 in rng<*p1,p2*> by TARSKI:def 2; hence <*p1,p2*>-:p2 = (<*p1,p2*> -| p2)^<*p2*> by Th44 .= <*p1*>^<*p2*> by A1,Th30 .= <*p1,p2*> by FINSEQ_1:def 9; end; theorem Th54: p1 <> p2 implies <*p1,p2,p3*>-:p2 = <*p1,p2*> proof assume A1: p1 <> p2; rng<*p1,p2,p3*> = { p1,p2,p3 } by Lm2; then p2 in rng<*p1,p2,p3*> by ENUMSET1:14; hence <*p1,p2,p3*>-:p2 = (<*p1,p2,p3*> -| p2)^<*p2*> by Th44 .= <*p1*>^<*p2*> by A1,Th31 .= <*p1,p2*> by FINSEQ_1:def 9; end; theorem Th55: p1 <> p3 & p2 <> p3 implies <*p1,p2,p3*>-:p3 = <*p1,p2,p3*> proof assume A1: p1 <> p3 & p2 <> p3; rng<*p1,p2,p3*> = { p1,p2,p3 } by Lm2; then p3 in rng<*p1,p2,p3*> by ENUMSET1:14; hence <*p1,p2,p3*>-:p3 = (<*p1,p2,p3*> -| p3)^<*p3*> by Th44 .= <*p1,p2*>^<*p3*> by A1,Th32 .= <*p1,p2,p3*> by FINSEQ_1:60; end; theorem <*p*>:-p = <*p*> & <*p*>-:p = <*p*> proof p in { p } by TARSKI:def 1; then A1: p in rng<*p*> by FINSEQ_1:56; hence <*p*>:-p = <*p*>^(<*p*>|--p )by Th45 .= <*p*>^{} by Th36 .= <*p*> by FINSEQ_1:47; thus <*p*>-:p = (<*p*>-|p)^<*p*> by A1,Th44 .= {}^<*p*> by Th36 .= <*p*> by FINSEQ_1:47; end; theorem Th57: p1 <> p2 implies <*p1,p2*>:-p2 = <*p2*> proof assume A1: p1 <> p2; p2 in { p1,p2 } by TARSKI:def 2; then p2 in rng<*p1,p2*> by Lm1; hence <*p1,p2*>:-p2 = <*p2*>^(<*p1,p2*> |-- p2) by Th45 .= <*p2*>^{} by A1,Th37 .= <*p2*> by FINSEQ_1:47; end; theorem Th58: p1 <> p2 implies <*p1,p2,p3*>:-p2 = <*p2,p3*> proof assume A1: p1 <> p2; p2 in { p1,p2,p3 } by ENUMSET1:14; then p2 in rng<*p1,p2,p3*> by Lm2; hence <*p1,p2,p3*>:-p2 = <*p2*>^(<*p1,p2,p3*> |-- p2) by Th45 .= <*p2*>^<*p3*> by A1,Th34 .= <*p2,p3*> by FINSEQ_1:def 9; end; theorem Th59: p1 <> p3 & p2 <> p3 implies <*p1,p2,p3*>:-p3 = <*p3*> proof assume A1: p1 <> p3 & p2 <> p3; p3 in { p1,p2,p3 } by ENUMSET1:14; then p3 in rng<*p1,p2,p3*> by Lm2; hence <*p1,p2,p3*>:-p3 = <*p3*>^(<*p1,p2,p3*> |-- p3) by Th45 .= <*p3*>^{} by A1,Th38 .= <*p3*> by FINSEQ_1:47; end; canceled; theorem Th61: p in rng f & p..f > k implies p..f = k + p..(f/^k) proof assume that A1: p in rng f and A2: p..f > k; reconsider i = p..f - k as Nat by A2,INT_1:18; A3: k+i = p..f by XCMPLX_1:27; i <> 0 by A2,XCMPLX_1:15; then A4: 1 <= i by RLVECT_1:99; A5: i + k <= len f by A1,A3,FINSEQ_4:31; then A6: i <= len f - k by REAL_1:84; k <= k + i by NAT_1:29; then k <= len f by A5,AXIOMS:22; then A7: i <= len(f/^k) by A6,RFINSEQ:def 2; then A8: i in dom(f/^k) by A4,FINSEQ_3:27; then A9: (f/^k)/.i = f/.(k+i) by FINSEQ_5:30 .= p by A1,A3,FINSEQ_5:41; now let j be Nat such that A10: 1 <= j and A11: j < i; j <= len(f/^k) by A7,A11,AXIOMS:22; then j in dom(f/^k) by A10,FINSEQ_3:27; then A12: f/.(k+j) = (f/^k)/.j by FINSEQ_5:30; A13: f/.(k+i) = (f/^k)/.i by A8,FINSEQ_5:30; A14: k+j < k+i by A11,REAL_1:53; k+j >= j & k+i <= len f by A1,A3,FINSEQ_4:31,NAT_1:29; then 1 <= k+j & k+j <= len f by A10,A14,AXIOMS:22; then A15: k+j in dom f by FINSEQ_3:27; k+j < p..f by A3,A11,REAL_1:53; then f/.(k+j) <> p by A15,FINSEQ_5:42; hence (f/^k)/.j <> (f/^k)/.i by A1,A3,A12,A13,FINSEQ_5:41; end; then p..f - k = p..(f/^k) by A8,A9,Th52; hence p..f = k+p..(f/^k) by XCMPLX_1:27; end; theorem Th62: p in rng f & p..f > k implies p in rng(f/^k) proof assume that A1: p in rng f and A2: p..f > k; p..f <= len f by A1,FINSEQ_4:31; then A3: k <= len f by A2,AXIOMS:22; A4: k+p..(f/^k) = p..f by A1,A2,Th61; then p..(f/^k) <> 0 by A2; then A5: 1 <= p..(f/^k) by RLVECT_1:99; len(f/^k) = len f - k by A3,RFINSEQ:def 2; then A6: len(f/^k) + k = len f by XCMPLX_1:27; k + p..(f/^k) <= len f by A1,A4,FINSEQ_4:31; then p..(f/^k) <= len(f/^k) by A6,REAL_1:53; then A7: p..(f/^k) in dom(f/^k) by A5,FINSEQ_3:27; then (f/^k)/.(p..(f/^k)) in rng(f/^k) by PARTFUN2:4; then f/.(k+p..(f/^k)) in rng(f/^k) by A7,FINSEQ_5:30; hence thesis by A1,A4,FINSEQ_5:41; end; theorem Th63: k < i & i in dom f implies f/.i in rng(f/^k) proof assume that A1: k < i and A2: i in dom f; reconsider j = i - k as Nat by A1,INT_1:18; A3: i = j + k by XCMPLX_1:27; j > 0 by A1,SQUARE_1:11; then A4: 1 <= j by RLVECT_1:99; A5: i <= len f by A2,FINSEQ_3:27; then k <= len f by A1,AXIOMS:22; then len(f/^k) = len f - k by RFINSEQ:def 2; then len(f/^k) + k = len f by XCMPLX_1:27; then j <= len(f/^k) by A3,A5,REAL_1:53; then A6: j in dom(f/^k) by A4,FINSEQ_3:27; then f/.i = (f/^k)/.j by A3,FINSEQ_5:30; hence thesis by A6,PARTFUN2:4; end; theorem Th64: p in rng f & p..f > k implies (f/^k)-:p = (f-:p)/^k proof assume that A1: p in rng f and A2: p..f > k; A3: k <= len(f-:p) by A1,A2,FINSEQ_5:45; A4: p..f = k + p..(f/^k) by A1,A2,Th61; A5: p in rng(f/^k) by A1,A2,Th62; then A6: len((f/^k)-:p) = p..(f/^k) by FINSEQ_5:45 .= p..f - k by A4,XCMPLX_1:26 .= len(f-:p) - k by A1,FINSEQ_5:45; now let m be Nat; assume A7: m in dom((f/^k)-:p); A8: m+k >= m by NAT_1:29; 1 <= m & m <= len(f-:p) - k by A6,A7,FINSEQ_3:27; then 1 <= m+k & m+k <= len(f-:p) by A8,AXIOMS:22,REAL_1:84; then A9: m+k in dom(f-:p) by FINSEQ_3:27; len(f-:p) = p..f by A1,FINSEQ_5:45; then A10: m+k in Seg(p..f) by A9,FINSEQ_1:def 3; len((f/^k)-:p) = p..(f/^k) by A5,FINSEQ_5:45; then A11: m in Seg(p..(f/^k)) by A7,FINSEQ_1:def 3; (f/^k)-:p = (f/^k)|(p..(f/^k)) by FINSEQ_5:def 1; then A12: dom((f/^k)-:p) c= dom(f/^k) by FINSEQ_5:20; thus ((f/^k)-:p).m = ((f/^k)-:p)/.m by A7,FINSEQ_4:def 4 .= (f/^k)/.m by A5,A11,FINSEQ_5:46 .= f/.(m+k) by A7,A12,FINSEQ_5:30 .= (f-:p)/.(m+k) by A1,A10,FINSEQ_5:46 .= (f-:p).(m+k) by A9,FINSEQ_4:def 4; end; hence (f/^k)-:p = (f-:p)/^k by A3,A6,RFINSEQ:def 2; end; theorem Th65: p in rng f & p..f <> 1 implies (f/^1)-:p = (f-:p)/^1 proof assume that A1: p in rng f and A2: p..f <> 1; p..f >= 1 by A1,FINSEQ_4:31; then p..f > 1 by A2,AXIOMS:21; hence (f/^1)-:p = (f-:p)/^1 by A1,Th64; end; theorem Th66: p in rng(f:-p) proof f:-p = <*p*>^(f/^p..f) by FINSEQ_5:def 2; then A1: rng(f:-p) = rng<*p*> \/ rng(f/^p..f) by FINSEQ_1:44; rng<*p*> = {p} by FINSEQ_1:56; then p in rng<*p*> by TARSKI:def 1; hence thesis by A1,XBOOLE_0:def 2; end; theorem Th67: x in rng f & p in rng f & x..f >= p..f implies x in rng(f:-p) proof assume that A1: x in rng f and A2: p in rng f and A3: x..f >= p..f; per cases by A3,AXIOMS:21; suppose A4: x..f > p..f; f:-p = <*p*>^(f/^p..f) by FINSEQ_5:def 2; then A5: rng(f:-p) = rng<*p*> \/ rng(f/^p..f) by FINSEQ_1:44; rng f c= D by FINSEQ_1:def 4; then reconsider q = x as Element of D by A1; q in rng(f/^(p..f)) by A1,A4,Th62; hence x in rng(f:-p) by A5,XBOOLE_0:def 2; suppose x..f = p..f; then x = f.(p..f) by A1,FINSEQ_4:29 .= p by A2,FINSEQ_4:29; hence x in rng(f:-p) by Th66; end; theorem Th68: p in rng f & k <= len f & k >= p..f implies f/.k in rng(f:-p) proof assume that A1: p in rng f and A2: k <= len f and A3: k >= p..f; set x = f/.k; per cases by A3,AXIOMS:21; suppose A4: k > p..f; f:-p = <*p*>^(f/^p..f) by FINSEQ_5:def 2; then A5: rng(f:-p) = rng<*p*> \/ rng(f/^p..f) by FINSEQ_1:44; reconsider q = x as Element of D; 1 <= p..f by A1,FINSEQ_4:31; then 1 <= k by A3,AXIOMS:22; then k in dom f by A2,FINSEQ_3:27; then q in rng(f/^(p..f)) by A4,Th63; hence x in rng(f:-p) by A5,XBOOLE_0:def 2; suppose k = p..f; then x = p by A1,FINSEQ_5:41; hence x in rng(f:-p) by Th66; end; theorem Th69: p in rng f1 implies (f1^f2):-p = (f1:-p)^f2 proof assume A1: p in rng f1; rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:44; then p in rng(f1^f2) by A1,XBOOLE_0:def 2; hence (f1^f2):-p = <*p*>^((f1^f2)|--p) by Th45 .= <*p*>^((f1|--p)^f2) by A1,Th10 .= <*p*>^(f1|--p)^f2 by FINSEQ_1:45 .= (f1:-p)^f2 by A1,Th45; end; theorem Th70: p in rng f2 \ rng f1 implies (f1^f2):-p = f2:-p proof assume A1: p in rng f2 \ rng f1; then A2: p in rng f2 by XBOOLE_0:def 4; rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:44; then p in rng(f1^f2) by A2,XBOOLE_0:def 2; hence (f1^f2):-p = <*p*>^((f1^f2)|--p) by Th45 .= <*p*>^(f2|--p) by A1,Th11 .= f2:-p by A2,Th45; end; theorem Th71: p in rng f1 implies (f1^f2)-:p = f1-:p proof assume A1: p in rng f1; rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:44; then p in rng(f1^f2) by A1,XBOOLE_0:def 2; hence (f1^f2)-:p = ((f1^f2)-|p)^<*p*> by Th44 .= (f1-|p)^<*p*> by A1,Th14 .= f1-:p by A1,Th44; end; theorem Th72: p in rng f2 \ rng f1 implies (f1^f2)-:p = f1^(f2-:p) proof assume A1: p in rng f2 \ rng f1; then A2: p in rng f2 by XBOOLE_0:def 4; rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:44; then p in rng(f1^f2) by A2,XBOOLE_0:def 2; hence (f1^f2)-:p = ((f1^f2)-|p)^<*p*> by Th44 .= f1^(f2-|p)^<*p*> by A1,Th17 .= f1^((f2-|p)^<*p*>) by FINSEQ_1:45 .= f1^(f2-:p) by A2,Th44; end; theorem f:-p:-p = f:-p proof A1: (<*p*>^(f/^p..f))/.1 = p by FINSEQ_5:16; thus f:-p:-p = (<*p*>^(f/^p..f)):-p by FINSEQ_5:def 2 .= <*p*>^(f/^p..f) by A1,Th48 .= f:-p by FINSEQ_5:def 2; end; theorem Th74: p1 in rng f & p2 in rng f \ rng(f-:p1) implies f|--p2 = f|--p1|--p2 proof assume that A1: p1 in rng f and A2: p2 in rng f \ rng(f-:p1); not p2 in rng(f-:p1) by A2,XBOOLE_0:def 4; then A3: not p2 in rng((f-|p1)^<* p1 *>) by A1,Th44; f = (f-|p1)^<* p1 *>^(f|--p1) by A1,FINSEQ_4:66; then A4: rng f = rng((f-|p1)^<* p1 *>) \/ rng(f|--p1) by FINSEQ_1:44; p2 in rng f by A2,XBOOLE_0:def 4; then p2 in rng(f|--p1) by A3,A4,XBOOLE_0:def 2; then A5: p2 in rng(f|--p1) \ rng((f-|p1)^<* p1 *>) by A3,XBOOLE_0:def 4; thus f|--p2 = ((f-|p1)^<* p1 *>^(f|--p1))|--p2 by A1,FINSEQ_4:66 .= f|--p1|--p2 by A5,Th11; end; theorem Th75: p in rng f implies rng f = rng(f-:p) \/ rng(f:-p) proof assume A1: p in rng f; then A2: f = (f-|p)^<* p *>^(f|--p) by FINSEQ_4:66; f-:p = (f -| p)^<*p*> by A1,Th44; then A3: rng( f-:p) = rng(f -| p) \/ rng<*p*> by FINSEQ_1:44; f:-p = <*p*>^(f |-- p) by A1,Th45; then A4: rng(f:-p) = rng<*p*> \/ rng(f |-- p) by FINSEQ_1:44; thus rng f = rng((f-|p)^<* p *>) \/ rng(f|--p) by A2,FINSEQ_1:44 .= rng(f-|p) \/ (rng<* p *> \/ rng<* p *>) \/ rng(f|--p) by FINSEQ_1:44 .= rng(f-|p) \/ rng<* p *> \/ rng<* p *> \/ rng(f|--p) by XBOOLE_1:4 .= rng(f-:p) \/ rng(f:-p) by A3,A4,XBOOLE_1:4; end; theorem Th76: p1 in rng f & p2 in rng f \ rng(f-:p1) implies f:-p1:-p2 = f:-p2 proof assume that A1: p1 in rng f and A2: p2 in rng f \ rng(f-:p1); A3: p2 in rng f by A2,XBOOLE_0:def 4; A4: rng f = rng(f-:p1) \/ rng(f:-p1) by A1,Th75; A5: not p2 in rng(f-:p1) by A2,XBOOLE_0:def 4; then A6: p2 in rng(f:-p1) by A3,A4,XBOOLE_0:def 2; f:-p1 = <*p1*>^(f |-- p1) by A1,Th45; then A7: rng(f:-p1) = rng<*p1*> \/ rng(f |-- p1) by FINSEQ_1:44; f-:p1 = (f -| p1)^<*p1*> by A1,Th44; then rng(f-:p1) = rng(f -| p1) \/ rng<*p1*> by FINSEQ_1:44; then A8: not p2 in rng<*p1*> by A5,XBOOLE_0:def 2; then p2 in rng(f|--p1) by A6,A7,XBOOLE_0:def 2; then A9: p2 in rng(f|--p1) \ rng<*p1*> by A8,XBOOLE_0:def 4; thus f:-p1:-p2 = <*p2*>^((f:-p1)|--p2) by A6,Th45 .= <*p2*>^((<*p1*>^(f|--p1))|--p2) by A1,Th45 .= <*p2*>^(f|--p1|--p2) by A9,Th11 .= <*p2*>^(f|--p2) by A1,A2,Th74 .= f:-p2 by A3,Th45; end; theorem Th77: p in rng f implies p..(f-:p) = p..f proof assume A1: p in rng f; then A2: (f-:p)/.(p..f) = p by FINSEQ_5:48; A3: 1 <= p..f by A1,FINSEQ_4:31; p..f <= len(f-:p) by A1,FINSEQ_5:45; then A4: p..f in dom(f-:p) by A3,FINSEQ_3:27; now let i such that A5: 1 <= i and A6: i < p..f; i in Seg(p..f) by A5,A6,FINSEQ_1:3; then A7: (f-:p)/.i = f/.i by A1,FINSEQ_5:46; p..f <> 0 by A1,FINSEQ_4:31; then p..f in Seg(p..f) by FINSEQ_1:5; then A8: (f-:p)/.(p..f) = f/.(p..f) by A1,FINSEQ_5:46; A9: f/.(p..f) = p by A1,FINSEQ_5:41; p..f <= len f by A1,FINSEQ_4:31; then i <= len f by A6,AXIOMS:22; then i in dom f by A5,FINSEQ_3:27; hence (f-:p)/.i <> (f-:p)/.(p..f) by A6,A7,A8,A9,FINSEQ_5:42; end; hence p..(f-:p) = p..f by A2,A4,Th52; end; theorem Th78: f|i|i = f|i proof f|i = f|Seg i by TOPREAL1:def 1; hence f|i|i = f|(Seg i)|Seg i by TOPREAL1:def 1 .= f|((Seg i) /\ Seg i) by RELAT_1:100 .= f|i by TOPREAL1:def 1; end; theorem Th79: p in rng f implies f-:p-:p = f-:p proof assume p in rng f; then A1: p..(f-:p) = p..f by Th77; thus f-:p-:p = (f-:p)|(p..(f-:p)) by FINSEQ_5:def 1 .= (f|(p..f))|(p..f) by A1,FINSEQ_5:def 1 .= f|(p..f) by Th78 .= f-:p by FINSEQ_5:def 1; end; theorem Th80: p1 in rng f & p2 in rng(f-:p1) implies f-:p1-:p2 = f-:p2 proof assume that A1: p1 in rng f and A2: p2 in rng(f-:p1); per cases; suppose p1 = p2; hence thesis by A1,Th79; suppose p1 <> p2; then not p2 in { p1 } by TARSKI:def 1; then A3: not p2 in rng<*p1*> by FINSEQ_1:56; f-:p1 = (f-|p1)^<*p1*> by A1,Th44; then rng(f-:p1) = rng(f-|p1) \/ rng<*p1*> by FINSEQ_1:44; then A4: p2 in rng(f-|p1) by A2,A3,XBOOLE_0:def 2; A5: rng(f-|p1) c= rng f by A1,FINSEQ_4:51; thus f-:p1-:p2 = ((f-:p1)-|p2)^<*p2*> by A2,Th44 .= (((f-|p1)^<*p1*>)-|p2)^<*p2*> by A1,Th44 .= ((f-|p1)-|p2)^<*p2*> by A4,Th14 .= (f-|p2)^<*p2*> by A1,A4,Th39 .= f-:p2 by A4,A5,Th44; end; theorem Th81: p in rng f implies (f-:p)^((f:-p)/^1) = f proof assume A1: p in rng f; then A2: rng(f|--p) c= rng f by FINSEQ_4:59; rng f c= D by FINSEQ_1:def 4; then rng(f|--p) c= D by A2,XBOOLE_1:1; then reconsider f1 = f|--p as FinSequence of D by FINSEQ_1:def 4; thus (f-:p)^((f:-p)/^1) = (f-|p)^<*p*>^((f:-p)/^1) by A1,Th44 .= (f-|p)^<*p*>^((<*p*>^f1)/^1) by A1,Th45 .= (f-|p)^<*p*>^(f|--p) by Th49 .= f by A1,FINSEQ_4:66; end; theorem Th82: f1 <> {} implies (f1^f2)/^1 = (f1/^1)^f2 proof assume f1 <> {}; then consider p being Element of D, df1 being FinSequence of D such that p = f1.1 and A1: f1 = <*p*>^df1 by FSM_1:5; thus (f1^f2)/^1 = (<*p*>^(df1^f2))/^1 by A1,FINSEQ_1:45 .= df1^f2 by Th49 .= (f1/^1)^f2 by A1,Th49; end; theorem Th83: p2 in rng f & p2..f <> 1 implies p2 in rng(f/^1) proof assume that A1: p2 in rng f and A2: p2..f <> 1; f = <*f/.1*>^(f/^1) by A1,FINSEQ_5:32,RELAT_1:60; then A3: rng f = rng<*f/.1*> \/ rng(f/^1) by FINSEQ_1:44; assume not p2 in rng(f/^1); then p2 in rng<*f/.1*> by A1,A3,XBOOLE_0:def 2; then p2 in { f/.1 } by FINSEQ_1:56; then p2 = f/.1 by TARSKI:def 1; hence contradiction by A1,A2,Th47,RELAT_1:60; end; theorem Th84: p..(f:-p) = 1 proof (f:-p)/.1 = p by FINSEQ_5:56; hence p..(f:-p) = 1 by Th47; end; Lm3: p in rng f & p..f > i implies i + p..(f/^i) = p..f proof assume that A1: p in rng f and A2: p..f > i; reconsider k = p..f - i as Nat by A2,INT_1:18; k <> 0 by A2,XCMPLX_1:15; then A3: 1 <= k by RLVECT_1:99; A4: p..f <= len f by A1,FINSEQ_4:31; then A5: i <= len f by A2,AXIOMS:22; p..f - i <= len f - i by A4,REAL_1:49; then A6: k <= len(f/^i) by A5,RFINSEQ:def 2; then A7: k in dom(f/^i) by A3,FINSEQ_3:27; then A8: (f/^i).k = f.(k+i) by A5,RFINSEQ:def 2 .= f.(p..f) by XCMPLX_1:27 .= p by A1,FINSEQ_4:29; now let j be Nat such that A9: 1 <= j and A10: j < k; j <= len(f/^i) by A6,A10,AXIOMS:22; then j in dom(f/^i) by A9,FINSEQ_3:27; then A11: f.(i+j) = (f/^i).j by A5,RFINSEQ:def 2; A12: f.(i+k) = (f/^i).k by A5,A7,RFINSEQ:def 2; A13: i + k = p..f by XCMPLX_1:27; then A14: i + j < p..f by A10,REAL_1:53; j <= i + j by NAT_1:29; then A15: 1 <= i + j by A9,AXIOMS:22; i + j <= len f by A4,A14,AXIOMS:22; then A16: i + j in dom f by A15,FINSEQ_3:27; f.(i+k) = p by A1,A13,FINSEQ_4:29; hence (f/^i).j <> (f/^i).k by A11,A12, A14,A16,FINSEQ_4:34; end; then p..(f/^i) = k by A7,A8,Th4; hence i + p..(f/^i) = p..f by XCMPLX_1:27; end; canceled; theorem Th86: <*>D/^k = <*>D proof per cases by NAT_1:19; suppose k = 0; hence thesis by FINSEQ_5:31; suppose k > 0; then k > len <*>D by FINSEQ_1:32; hence thesis by RFINSEQ:def 2; end; theorem Th87: f/^(i+k) = f/^i/^k proof per cases; suppose A1: i+k <= len f; i <= i+k by NAT_1:29; then A2: i <= len f by A1,AXIOMS:22; then A3: len(f/^i) = len f - i by RFINSEQ:def 2; then A4: k <= len(f/^i) by A1,REAL_1:84; then A5: len(f/^i/^k) = len(f/^i) - k by RFINSEQ:def 2 .= len f -(i+k) by A3,XCMPLX_1:36; now let m be Nat; assume A6: m in dom(f/^i/^k); then A7: m+k in dom(f/^i) by FINSEQ_5:29; thus (f/^i/^k).m = (f/^i).(m+k) by A4,A6,RFINSEQ:def 2 .= f.(m+k+i) by A2,A7,RFINSEQ:def 2 .= f.(m+(i+k)) by XCMPLX_1:1; end; hence f/^(i+k) = f/^i/^k by A1,A5,RFINSEQ:def 2; suppose that A8: i+k > len f and A9: i<= len f; len(f/^i) = len f - i by A9,RFINSEQ:def 2; then len(f/^i) + i = len f by XCMPLX_1:27; then A10: k > len(f/^i) by A8,AXIOMS:24; thus f/^(i+k) = <*>D by A8,RFINSEQ:def 2 .= f/^i/^k by A10,RFINSEQ:def 2; suppose that A11: i+k > len f and A12: i > len f; thus f/^(i+k) = <*>D by A11,RFINSEQ:def 2 .= <*>D/^k by Th86 .= f/^i/^k by A12,RFINSEQ:def 2; end; theorem Th88: p in rng f & p..f > k implies (f/^k):-p = f:-p proof assume A1: p in rng f & p..f > k; thus (f/^k):-p =<*p*>^(f/^k/^(p..(f/^k))) by FINSEQ_5:def 2 .=<*p*>^(f/^(k + p..(f/^k))) by Th87 .=<*p*>^(f/^p..f) by A1,Lm3 .= f:-p by FINSEQ_5:def 2; end; theorem Th89: p in rng f & p..f <> 1 implies (f/^1):-p = f:-p proof assume that A1: p in rng f and A2: p..f <> 1; p..f >= 1 by A1,FINSEQ_4:31; then p..f > 1 by A2,AXIOMS:21; hence thesis by A1,Th88; end; theorem Th90: i + k = len f implies Rev(f/^k) = Rev f|i proof assume A1: i + k = len f; then A2: i <= len f by NAT_1:29; A3: k <= len f by A1,NAT_1:29; i <= len Rev f by A2,FINSEQ_5:def 3; then A4: len(Rev f|i) = i by TOPREAL1:3 .= len f - k by A1,XCMPLX_1:26 .= len(f/^k) by A3,RFINSEQ:def 2; now let j be Nat; assume A5: j in dom(Rev f|i); A6: dom(Rev f|i) c= dom Rev f by FINSEQ_5:20; A7: len(f/^k) = len f - k by A3,RFINSEQ:def 2; j <= len(f/^k) by A4,A5,FINSEQ_3:27; then reconsider m = len(f/^k) - j as Nat by INT_1:18; A8: 1 <= m + 1 by NAT_1:29; j >= 1 by A5,FINSEQ_3:27; then len(f/^k) - j <= len(f/^k) - 1 by REAL_2:106; then len(f/^k) - j + 1 <= len(f/^k) by REAL_1:84; then A9: m + 1 in dom(f/^k) by A8,FINSEQ_3:27; thus (Rev f|i).j = (Rev f|i)/.j by A5,FINSEQ_4:def 4 .= (Rev f)/.j by A5,TOPREAL1:1 .= (Rev f).j by A5,A6,FINSEQ_4:def 4 .= f.(len f - j + 1) by A5,A6,FINSEQ_5:def 3 .= f.(len(f/^k) + k - j + 1) by A7,XCMPLX_1:27 .= f.(m + k + 1) by XCMPLX_1:29 .= f.(m + 1 + k) by XCMPLX_1:1 .= (f/^k).(len(f/^k) - j + 1) by A3,A9,RFINSEQ:def 2; end; hence Rev(f/^k) = Rev f|i by A4,FINSEQ_5:def 3; end; theorem Th91: i + k = len f implies Rev(f|k) = Rev f/^i proof assume i + k = len f; then A1: i + k = len Rev f by FINSEQ_5:def 3; thus Rev(f|k) = Rev(Rev Rev f |k) by Th29 .= Rev Rev(Rev f/^i) by A1,Th90 .= Rev f/^i by Th29; end; theorem Th92: f just_once_values p implies Rev(f|--p) = Rev f -|p proof assume A1: f just_once_values p; then A2: p in rng f by FINSEQ_4:7; then A3: p in rng Rev f by FINSEQ_5:60; then p..Rev f >= 1 by FINSEQ_4:31; then reconsider n = p..Rev f - 1 as Nat by INT_1:18; p..f + p..Rev f = len f + 1 by A1,Th41; then p..f + p..Rev f -1 = len f by XCMPLX_1:26; then A4: n + p..f = len f by XCMPLX_1:29; Rev(f|--p) = Rev(f/^(p..f)) by A2,FINSEQ_5:38 .= (Rev f) | n by A4,Th90 .= (Rev f) | Seg n by TOPREAL1:def 1; hence Rev(f|--p) = Rev f -|p by A3,FINSEQ_4:def 6; end; theorem Th93: f just_once_values p implies Rev(f:-p) = Rev f -:p proof assume A1: f just_once_values p; then A2: p in rng f by FINSEQ_4:7; then A3: p in rng Rev f by FINSEQ_5:60; thus Rev(f:-p) = Rev(<*p*>^(f|--p)) by A2,Th45 .= Rev(f|--p)^<*p*> by Th28 .= (Rev f -|p)^<*p*> by A1,Th92 .= Rev f -:p by A3,Th44; end; theorem Th94: f just_once_values p implies Rev(f-:p) = Rev f :-p proof assume A1: f just_once_values p; then A2: p in rng f by FINSEQ_4:7; then A3: p in rng Rev f by FINSEQ_5:60; thus Rev(f-:p) = Rev((f-|p)^<*p*>) by A2,Th44 .= <*p*>^Rev(f-|p) by FINSEQ_5:66 .= <*p*>^(Rev f |-- p) by A1,Th42 .= Rev f :-p by A3,Th45; end; begin :: rotation, circular definition let D be non empty set; let IT be FinSequence of D; attr IT is circular means :Def1: IT/.1 = IT/.len IT; end; definition let D,f,p; func Rotate(f,p) -> FinSequence of D equals :Def2: (f:-p)^((f-:p)/^1) if p in rng f otherwise f; correctness; end; definition let D; let f be non empty FinSequence of D, p be Element of D; cluster Rotate(f,p) -> non empty; coherence proof per cases; suppose A1: p in rng f; (f:-p)^((f-:p)/^1) is non empty; hence thesis by A1,Def2; suppose not p in rng f; hence thesis by Def2; end; end; definition let D; cluster circular non empty trivial FinSequence of D; existence proof consider d being Element of D; take <*d*>; thus <*d*> is circular proof thus <*d*>/.1 = <*d*>/.len<*d*> by FINSEQ_1:56; end; thus thesis; end; cluster circular non empty non trivial FinSequence of D; existence proof consider d being Element of D; take <*d,d*>; thus <*d,d*> is circular proof len<*d,d*> = 2 by FINSEQ_1:61; hence <*d,d*>/.len<*d,d*> = d by FINSEQ_4:26 .= <*d,d*>/.1 by FINSEQ_4:26; end; thus thesis; end; end; theorem Th95: Rotate(f,f/.1) = f proof A1: len<*f/.1*> = 1 by FINSEQ_1:56; per cases; suppose A2: f is non empty; then f/.1 in rng f by Th46; hence Rotate(f,f/.1) = (f:-f/.1)^((f-:f/.1)/^1) by Def2 .= f^((f-:f/.1)/^1) by A2,Th48 .= f^(<*f/.1*>/^1) by A2,Th48 .= f^{} by A1,FINSEQ_5:35 .= f by FINSEQ_1:47; suppose f is empty; hence thesis by Def2,RELAT_1:60; end; definition let D,p; let f be circular non empty FinSequence of D; cluster Rotate(f,p) -> circular; coherence proof per cases; suppose not p in rng f; hence thesis by Def2; suppose that A1: p in rng f and A2: p <> f/.1; A3: Rotate(f,p) = (f:-p)^((f-:p)/^1) by A1,Def2; p..f <> 1 & p..f >= 1 by A1,A2,FINSEQ_4:31,FINSEQ_5:41; then p..f > 1 by REAL_1:def 5; then p..f >= 1+1 by NAT_1:38; then A4: len(f-:p) >= 1+1 by A1,FINSEQ_5:45; then 1 <= len(f-:p) by AXIOMS:22; then len((f-:p)/^1) = len(f-:p)-1 by RFINSEQ:def 2; then A5: len((f-:p)/^1) + 1 = len(f-:p) by XCMPLX_1:27; then len((f-:p)/^1) >= 1 by A4,REAL_1:53; then A6: len((f-:p)/^1) in dom((f-:p)/^1) by FINSEQ_3:27; 1 in dom(f:-p) by FINSEQ_5:6; hence (Rotate(f,p))/.1 = (f:-p)/.1 by A3,GROUP_5:95 .= p by FINSEQ_5:56 .= (f-:p)/.(p..f) by A1,FINSEQ_5:48 .= (f-:p)/.(len((f-:p)/^1)+1) by A1,A5,FINSEQ_5:45 .= ((f-:p)/^1)/.(len((f-:p)/^1)) by A6,FINSEQ_5:30 .= ((f:-p)^((f-:p)/^1))/.(len(f:-p)+len((f-:p)/^1)) by A6,GROUP_5:96 .= (Rotate(f,p))/.len Rotate(f,p) by A3,FINSEQ_1:35; suppose p in rng f & p = f/.1; hence thesis by Th95; end; end; theorem f is circular & p in rng f implies rng Rotate(f,p) = rng f proof assume that A1: f is circular and A2: p in rng f; A3: Rotate(f,p) = (f:-p)^((f-:p)/^1) by A2,Def2; A4: rng((f:-p)^((f-:p)/^1)) = rng(f:-p) \/ rng((f-:p)/^1) by FINSEQ_1:44; A5: rng(f:-p) c= rng f by A2,FINSEQ_5:58; A6: rng(f-:p) c= rng f by FINSEQ_5:51; rng((f-:p)/^1) c= rng(f-:p) by FINSEQ_5:36; then rng((f-:p)/^1) c= rng f by A6,XBOOLE_1:1; hence rng Rotate(f,p) c= rng f by A3,A4,A5,XBOOLE_1:8; let x be set; assume x in rng f; then consider i such that A7: i in dom f and A8: f.i = x by FINSEQ_2:11; A9: x = f/.i by A7,A8,FINSEQ_4:def 4; per cases; suppose i = 1; then A10: x = f/.len f by A1,A9,Def1 .= (f:-p)/.len(f:-p) by A2,FINSEQ_5:57; len(f:-p) = len(<*p*>^(f/^p..f)) by FINSEQ_5:def 2 .= len<*p*> + len(f/^p..f) by FINSEQ_1:35 .= 1 + len(f/^p..f) by FINSEQ_1:56; then 1 <= len(f:-p) by NAT_1:29; then len(f:-p) in dom(f:-p) by FINSEQ_3:27; then x in rng(f:-p) by A10,PARTFUN2:4; hence x in rng Rotate(f,p) by A3,A4,XBOOLE_0:def 2; suppose that A11: i <= p..f and A12: i <> 1; A13: i <> 0 by A7,FINSEQ_3:27; then A14: i > 0+1 by A12,CQC_THE1:2; consider j being Nat such that A15: i = j+1 by A13,NAT_1:22; A16: i <= len(f-:p) by A2,A11,FINSEQ_5:45; then 1 <= len(f-:p) by A14,AXIOMS:22; then len((f-:p)/^1) = len(f-:p) - 1 by RFINSEQ:def 2; then len((f-:p)/^1) + 1 = len(f-:p) by XCMPLX_1:27; then A17: j <= len((f-:p)/^1) by A15,A16,REAL_1:53; j >= 1 by A14,A15,RLVECT_1:99; then A18: j in dom((f-:p)/^1) by A17,FINSEQ_3:27; A19: len<*(f-:p)/.1*> = 1 by FINSEQ_1:56; A20: i in Seg(p..f) by A11,A14,FINSEQ_1:3; f-:p is non empty by A2,FINSEQ_5:50; then f-:p = <*(f-:p)/.1*>^((f-:p)/^1) by FINSEQ_5:32; then ((f-:p)/^1)/.j = (f-:p)/.i by A15,A18,A19,GROUP_5:96 .= f/.i by A2,A20,FINSEQ_5:46; then x in rng((f-:p)/^1) by A9,A18,PARTFUN2:4; hence x in rng Rotate(f,p) by A3,A4,XBOOLE_0:def 2; suppose i > p..f; then reconsider j = i - p..f as Nat by INT_1:18; A21: j+1 >= 1 by NAT_1:29; A22: len (f:-p) = len f - p..f + 1 by A2,FINSEQ_5:53; i <= len f by A7,FINSEQ_3:27; then j <= len f - p..f by REAL_1:49; then j + 1 <= len(f:-p) by A22,AXIOMS:24; then A23: j+1 in dom(f:-p) by A21,FINSEQ_3:27; j + p..f = i by XCMPLX_1:27; then f/.i = (f:-p)/.(j+1) by A2,A23,FINSEQ_5:55; then x in rng(f:-p) by A9,A23,PARTFUN2:4; hence x in rng Rotate(f,p) by A3,A4,XBOOLE_0:def 2; end; theorem Th97: p in rng f implies p in rng Rotate(f,p) proof assume p in rng f; then Rotate(f,p) = (f:-p)^((f-:p)/^1) by Def2; then A1: rng Rotate(f,p) = rng(f:-p) \/ rng((f-:p)/^1) by FINSEQ_1:44; p in {p} by TARSKI:def 1; then p in rng<*p*> by FINSEQ_1:56; then p in rng<*p*> \/ rng(f/^p..f) by XBOOLE_0:def 2; then p in rng(<*p*>^(f/^p..f)) by FINSEQ_1:44; then p in rng(f:-p) by FINSEQ_5:def 2; hence p in rng Rotate(f,p) by A1,XBOOLE_0:def 2; end; theorem Th98: p in rng f implies (Rotate(f,p))/.1 = p proof assume p in rng f; then Rotate(f,p) =(f:-p)^((f-:p)/^1) by Def2 .= <*p*>^(f/^p..f)^((f-:p)/^1) by FINSEQ_5:def 2 .= <*p*>^((f/^p..f)^((f-:p)/^1)) by FINSEQ_1:45; hence thesis by FINSEQ_5:16; end; theorem Th99: Rotate(Rotate(f,p),p) = Rotate(f,p) proof per cases; suppose A1: p in rng f; then reconsider f' = f as non empty FinSequence of D by RELAT_1:60; (Rotate(f,p))/.1 = p by A1,Th98; then A2: Rotate(f',p)-:p = <*p*> & Rotate(f',p):-p = Rotate(f,p) by Th48; A3: len<*p*> = 1 by FINSEQ_1:56; p in rng Rotate(f,p) by A1,Th97; hence Rotate(Rotate(f,p),p) = Rotate(f,p)^(<*p*>/^1) by A2,Def2 .= Rotate(f,p)^{} by A3,FINSEQ_5:35 .= Rotate(f,p) by FINSEQ_1:47; suppose not p in rng f; hence thesis by Def2; end; theorem Rotate(<*p*>,p) = <*p*> proof <*p*>/.1 = p by FINSEQ_4:25; hence thesis by Th95; end; theorem Th101: Rotate(<*p1,p2*>,p1) = <*p1,p2*> proof <*p1,p2*>/.1 = p1 by FINSEQ_4:26; hence thesis by Th95; end; theorem Rotate(<*p1,p2*>,p2) = <*p2,p2*> proof per cases; suppose p1 = p2; hence thesis by Th101; suppose A1: p1 <> p2; rng<*p1,p2*> = {p1,p2} by Lm1; then p2 in rng<*p1,p2*> by TARSKI:def 2; hence Rotate(<*p1,p2*>,p2) = (<*p1,p2*>:-p2)^((<*p1,p2*>-:p2)/^1) by Def2 .= <*p2*>^((<*p1,p2*>-:p2)/^1) by A1,Th57 .= <*p2*>^(<*p1,p2*>/^1) by A1,Th53 .= <*p2*>^<*p2*> by Th50 .= <*p2,p2*> by FINSEQ_1:def 9; end; theorem Th103: Rotate(<*p1,p2,p3*>,p1) = <*p1,p2,p3*> proof <*p1,p2,p3*>/.1 = p1 by FINSEQ_4:27; hence thesis by Th95; end; theorem p1 <> p2 implies Rotate(<*p1,p2,p3*>,p2) = <*p2,p3,p2*> proof assume A1: p1 <> p2; rng<*p1,p2,p3*> = {p1,p2,p3} by Lm2; then p2 in rng<*p1,p2,p3*> by ENUMSET1:14; hence Rotate(<*p1,p2,p3*>,p2) = (<*p1,p2,p3*>:-p2)^((<*p1,p2,p3*>-:p2)/^1) by Def2 .= <*p2,p3*>^((<*p1,p2,p3*>-:p2)/^1) by A1,Th58 .= <*p2,p3*>^(<*p1,p2*>/^1) by A1,Th54 .= <*p2,p3*>^<*p2*> by Th50 .= <*p2,p3,p2*> by FINSEQ_1:60; end; theorem p2 <> p3 implies Rotate(<*p1,p2,p3*>,p3) = <*p3,p2,p3*> :: (-: proof assume A1: p2 <> p3; per cases; suppose p1 = p3; hence thesis by Th103; suppose A2: p1 <> p3; rng<*p1,p2,p3*> = {p1,p2,p3} by Lm2; then p3 in rng<*p1,p2,p3*> by ENUMSET1:14; hence Rotate(<*p1,p2,p3*>,p3) = (<*p1,p2,p3*>:-p3)^((<*p1,p2,p3*>-:p3)/^1) by Def2 .= <*p3*>^((<*p1,p2,p3*>-:p3)/^1) by A1,A2,Th59 .= <*p3*>^(<*p1,p2,p3*>/^1) by A1,A2,Th55 .= <*p3*>^<*p2,p3*> by Th51 .= <*p3,p2,p3*> by FINSEQ_1:60; end; theorem for f being circular non trivial FinSequence of D holds rng(f/^1) = rng f proof let f be circular non trivial FinSequence of D; thus rng(f/^1) c= rng f by FINSEQ_5:36; let x be set; f = <*f/.1*>^(f/^1) by FINSEQ_5:32; then A1: rng f = rng<*f/.1*> \/ rng(f/^1) by FINSEQ_1:44; assume A2: x in rng f; per cases by A1,A2,XBOOLE_0:def 2; suppose x in rng<*f/.1*>; then x in {f/.1} by FINSEQ_1:56; then x = f/.1 by TARSKI:def 1; then A3: x = f/.len f by Def1; A4: len f >= 1+1 by SPPOL_1:2; then len f >= 1 by AXIOMS:22; then A5: len(f/^1) = len f - 1 by RFINSEQ:def 2; then 1 <= len(f/^1) by A4,REAL_1:84; then A6: len(f/^1) in dom(f/^1) by FINSEQ_3:27; len(f/^1) + 1 = len f by A5,XCMPLX_1:27; then x = (f/^1)/.len(f/^1) by A3,A6,FINSEQ_5:30; hence x in rng(f/^1) by A6,PARTFUN2:4; suppose x in rng(f/^1); hence thesis; end; theorem Th107: rng(f/^1) c= rng Rotate(f,p) proof per cases; suppose A1: p in rng f; then Rotate(f,p) = (f:-p)^((f-:p)/^1) by Def2; then A2: rng Rotate(f,p) = rng(f:-p) \/ rng((f-:p)/^1) by FINSEQ_1:44; thus thesis proof let x be set; assume A3: x in rng(f/^1); A4: rng(f/^1) c= rng f by FINSEQ_5:36; then A5: x in rng f by A3; per cases; suppose that A6: x..f < p..f and A7: x..(f/^1) >= p..(f/^1); A8: p..f <> 1 by A3,A4,A6,FINSEQ_4:31; p..f >= 1 by A1,FINSEQ_4:31; then p..f > 1 by A8,AXIOMS:21; then A9: p..f = p..(f/^1) + 1 by A1,Th61; len f <> 0 by A1,FINSEQ_1:25,RELAT_1:60; then len f >= 1 by RLVECT_1:99; then len f - 1 = len(f/^1) by RFINSEQ:def 2; then A10: len f = len(f/^1) + 1 by XCMPLX_1:27; x..(f/^1) <= len(f/^1) by A3,FINSEQ_4:31; then A11: x..(f/^1)+1 <= len f by A10,AXIOMS:24; A12: x..(f/^1)+1 >= p..f by A7,A9,AXIOMS:24; A13: x..(f/^1) in dom(f/^1) by A3,FINSEQ_4:30; rng f c= D by FINSEQ_1:def 4; then reconsider q = x as Element of D by A5; q = (f/^1)/.(q..(f/^1)) by A3,FINSEQ_5:41 .= f/.(q..(f/^1)+1) by A13,FINSEQ_5:30; then x in rng(f:-p) by A1,A11,A12,Th68; hence thesis by A2,XBOOLE_0:def 2; suppose that A14: x..f < p..f and A15: x..(f/^1) <= p..(f/^1); A16: p..f <> 1 by A3,A4,A14,FINSEQ_4:31; p..f >= 1 by A1,FINSEQ_4:31; then p..f > 1 by A16,AXIOMS:21; then p in rng(f/^1) by A1,Th62; then x in rng((f/^1)-:p) by A3,A15,FINSEQ_5:49; then x in rng((f-:p)/^1) by A1,A16,Th65; hence thesis by A2,XBOOLE_0:def 2; suppose x..f >= p..f; then x in rng(f:-p) by A1,A3,A4,Th67; hence thesis by A2,XBOOLE_0:def 2; end; suppose not p in rng f; then Rotate(f,p) = f by Def2; hence thesis by FINSEQ_5:36; end; theorem Th108: p2 in rng f \ rng(f-:p1) implies Rotate(Rotate(f,p1),p2) = Rotate(f,p2) proof assume A1: p2 in rng f \ rng(f-:p1); per cases; suppose A2: p1 in rng f; then A3: Rotate(f,p1) = (f:-p1)^((f-:p1)/^1) by Def2; A4: p2 in rng f by A1,XBOOLE_0:def 4; A5: not p2 in rng(f-:p1) by A1,XBOOLE_0:def 4; rng f = rng(f-:p1) \/ rng(f:-p1) by A2,Th75; then A6: p2 in rng(f:-p1) by A4,A5,XBOOLE_0:def 2; A7: f-:p1 <> {} by A2,FINSEQ_5:50; rng((f-:p1)/^1) c= rng(f-:p1) by FINSEQ_5:36; then A8: not p2 in rng((f-:p1)/^1) by A1,XBOOLE_0:def 4; A9: p1 in rng(f:-p1) by Th66; p1..f <= p1..f; then A10: p1 <> p2 by A2,A5,FINSEQ_5:49; p1..(f:-p1) = 1 by Th84; then A11: p2..(f:-p1) <> 1 by A6,A9,A10,FINSEQ_5:10; A12: now assume p2..f = 1; then p2..f <= p1..f by A2,FINSEQ_4:31; hence contradiction by A2,A4,A5,FINSEQ_5:49; end; then A13: p2 in rng(f/^1) by A4,Th83; p2 in rng((f:-p1)/^1) by A6,A11,Th83; then A14: p2 in rng((f:-p1)/^1)\rng((f-:p1)/^1) by A8,XBOOLE_0:def 4; rng(f/^1) c= rng Rotate(f,p1) by Th107; hence Rotate(Rotate(f,p1),p2) = (((f:-p1)^((f-:p1)/^1)):-p2)^((((f:-p1)^((f-:p1)/^1))-:p2)/^1) by A3 ,A13,Def2 .= (f:-p1:-p2)^((f-:p1)/^1)^((((f:-p1)^((f-:p1)/^1))-:p2)/^1) by A6,Th69 .= (f:-p2)^((f-:p1)/^1)^((((f:-p1)^((f-:p1)/^1))-:p2)/^1) by A1,A2,Th76 .= (f:-p2)^((f-:p1)/^1)^(((f:-p1)-:p2)/^1) by A6,Th71 .= (f:-p2)^((f-:p1)/^1)^(((f:-p1)/^1)-:p2) by A6,A11,Th65 .= (f:-p2)^(((f-:p1)/^1)^(((f:-p1)/^1)-:p2)) by FINSEQ_1:45 .= (f:-p2)^((((f-:p1)/^1)^((f:-p1)/^1))-:p2) by A14,Th72 .= (f:-p2)^((((f-:p1)^((f:-p1)/^1))/^1)-:p2) by A7,Th82 .= (f:-p2)^((f/^1)-:p2) by A2,Th81 .= (f:-p2)^((f-:p2)/^1) by A4,A12,Th65 .= Rotate(f,p2) by A4,Def2; suppose not p1 in rng f; hence thesis by Def2; end; theorem Th109: p2..f <> 1 & p2 in rng f \ rng(f:-p1) implies Rotate(Rotate(f,p1),p2) = Rotate(f,p2) proof assume that A1: p2..f <> 1 and A2: p2 in rng f \ rng(f:-p1); per cases; suppose A3: p1 in rng f; then A4: Rotate(f,p1) = (f:-p1)^((f-:p1)/^1) by Def2; A5: not p2 in rng(f:-p1) by A2,XBOOLE_0:def 4; A6: p2 in rng f by A2,XBOOLE_0:def 4; rng f = rng(f-:p1) \/ rng(f:-p1) by A3,Th75; then A7: p2 in rng(f-:p1) by A5,A6,XBOOLE_0:def 2; A8: f-:p1 <> {} by A3,FINSEQ_5:50; (f-:p1)^((f:-p1)/^1) = f by A3,Th81; then A9: p2..(f-:p1) <> 1 by A1,A7,Th8; then A10: p2 in rng((f-:p1)/^1) by A7,Th83; then A11: p2 in rng((f-:p1)/^1) \ rng(f:-p1) by A5,XBOOLE_0:def 4; rng Rotate(f,p1) = rng(f:-p1) \/ rng((f-:p1)/^1) by A4,FINSEQ_1:44; then p2 in rng Rotate(f,p1) by A10,XBOOLE_0:def 2; hence Rotate(Rotate(f,p1),p2) = (((f:-p1)^((f-:p1)/^1)):-p2)^((((f:-p1)^((f-:p1)/^1))-:p2)/^1) by A4 ,Def2 .= (((f-:p1)/^1):-p2)^((((f:-p1)^((f-:p1)/^1))-:p2)/^1) by A11,Th70 .= (((f-:p1)/^1):-p2)^(((f:-p1)^(((f-:p1)/^1)-:p2))/^1) by A11,Th72 .= (((f-:p1)/^1):-p2)^(((f:-p1)/^1)^(((f-:p1)/^1)-:p2)) by Th82 .= (((f-:p1)/^1):-p2)^((f:-p1)/^1)^(((f-:p1)/^1)-:p2) by FINSEQ_1:45 .= ((((f-:p1)/^1)^((f:-p1)/^1)):-p2)^(((f-:p1)/^1)-:p2) by A10,Th69 .= ((((f-:p1)^((f:-p1)/^1)/^1)):-p2)^(((f-:p1)/^1)-:p2) by A8,Th82 .= ((f/^1):-p2)^(((f-:p1)/^1)-:p2) by A3,Th81 .= (f:-p2)^(((f-:p1)/^1)-:p2) by A1,A6,Th89 .= (f:-p2)^((f-:p1-:p2)/^1) by A7,A9,Th65 .= (f:-p2)^((f-:p2)/^1) by A3,A7,Th80 .= Rotate(f,p2) by A6,Def2; suppose not p1 in rng f; hence thesis by Def2; end; theorem Th110: p2 in rng(f/^1) & f just_once_values p2 implies Rotate(Rotate(f,p1),p2) = Rotate(f,p2) proof per cases; suppose A1: p1 in rng f; assume that A2: p2 in rng(f/^1) and A3: f just_once_values p2; A4: p2 in rng f by A3,FINSEQ_4:7; f = (f -| p1) ^ <* p1 *> ^ (f |-- p1) by A1,FINSEQ_4:66; then A5: p2 in rng((f -| p1)^<* p1 *>) \+\ rng(f |-- p1) by A3,Th19; A6: now per cases by A5,XBOOLE_0:1; case that p2 in rng((f -| p1)^<* p1 *>) and A7: not p2 in rng(f |-- p1); now per cases; case not p2 in rng<* p1 *>; then not p2 in rng(f |-- p1) \/ rng<* p1 *> by A7,XBOOLE_0:def 2; then not p2 in rng(<* p1 *>^(f |-- p1)) by FINSEQ_1:44; hence not p2 in rng(f:-p1) by A1,Th45; case p2 in rng<* p1 *>; then p2 in { p1 } by FINSEQ_1:56; hence p2 = p1 by TARSKI:def 1; end; hence p2 in rng(f:-p1) implies p1 = p2; case not p2 in rng((f -| p1)^<* p1 *>); hence not p2 in rng(f-:p1) by A1,Th44; end; f = <*f/.1*>^(f/^1) by A4,FINSEQ_5:32,RELAT_1:60; then p2 in rng<*f/.1*> \+\ rng(f/^1) by A3,Th19; then not p2 in rng<*f/.1*> by A2,XBOOLE_0:1; then not p2 in {f/.1} by FINSEQ_1:56; then p2 <> f/.1 by TARSKI:def 1; then A8: p2..f <> 1 by A4,FINSEQ_5:41; now per cases by A4,A6,XBOOLE_0:def 4; suppose p1 = p2; hence Rotate(Rotate(f,p1),p2) = Rotate(f,p2) by Th99; suppose p2 in rng f \ rng(f-:p1); hence Rotate(Rotate(f,p1),p2) = Rotate(f,p2) by Th108; suppose p2 in rng f \ rng(f:-p1); hence Rotate(Rotate(f,p1),p2) = Rotate(f,p2) by A8,Th109; end; hence thesis; suppose not p1 in rng f; hence thesis by Def2; end; theorem f is circular & f just_once_values p2 implies Rotate(Rotate(f,p1),p2) = Rotate(f,p2) proof per cases; suppose A1: p1 in rng f; assume that A2: f is circular and A3: f just_once_values p2; A4: p2 in rng f by A3,FINSEQ_4:60; now per cases; suppose rng f is trivial; then p1 = p2 by A1,A4,SPPOL_1:3; hence Rotate(Rotate(f,p1),p2) = Rotate(f,p2) by Th99; suppose A5: rng f is not trivial; then f = <*f/.1*>^(f/^1) by FINSEQ_5:32,RELAT_1:60; then A6: rng f = rng<*f/.1*> \/ rng(f/^1) by FINSEQ_1:44; now assume A7: not p2 in rng(f/^1); then p2 in rng<*f/.1*> by A4,A6,XBOOLE_0:def 2; then p2 in {f/.1} by FINSEQ_1:56; then p2 = f/.1 by TARSKI:def 1; then A8: p2 = f/.len f by A2,Def1; f qua set is non trivial by A5,Th21; then len f >= 1 + 1 by SPPOL_1:2; then A9: 1 < len f by NAT_1:38; len f in dom f by A5,FINSEQ_5:6,RELAT_1:60; hence contradiction by A7,A8,A9,Th63; end; hence Rotate(Rotate(f,p1),p2) = Rotate(f,p2) by A3,Th110; end; hence Rotate(Rotate(f,p1),p2) = Rotate(f,p2); suppose not p1 in rng f; hence thesis by Def2; end; theorem f is circular & f just_once_values p implies Rev Rotate(f,p) = Rotate(Rev f,p) proof assume that A1: f is circular and A2: f just_once_values p; A3: p in rng f by A2,FINSEQ_4:60; then A4: p in rng Rev f by FINSEQ_5:60; A5: f/.1 = f/.len f by A1,Def1; 0 <> len(f:-p) by FINSEQ_1:25; then 1 <= len(f:-p) by RLVECT_1:99; then reconsider j = len(f:-p) - 1 as Nat by INT_1:18; A6: j + 1 = len(f:-p) by XCMPLX_1:27; f-:p is non empty by A3,FINSEQ_5:50; then A7: f-:p = <*(f-:p)/.1*>^((f-:p)/^1) by FINSEQ_5:32 .= <*f/.1*>^((f-:p)/^1) by A3,FINSEQ_5:47; A8: f:-p = ((f:-p)|j)^<*(f:-p)/.len(f:-p)*> by A6,FINSEQ_5:24 .= ((f:-p)|j)^<*f/.len f*> by A3,FINSEQ_5:57; thus Rev Rotate(f,p) = Rev((f:-p)^((f-:p)/^1)) by A3,Def2 .= Rev((f-:p)/^1)^Rev(f:-p) by FINSEQ_5:67 .= Rev((f-:p)/^1)^(<*f/.len f*>^Rev((f:-p)|j)) by A8,FINSEQ_5:66 .= Rev((f-:p)/^1)^<*f/.1*>^Rev((f:-p)|j) by A5,FINSEQ_1:45 .= Rev(f-:p)^Rev((f:-p)|j) by A7,Th28 .= Rev(f-:p)^(Rev(f:-p)/^1) by A6,Th91 .= Rev(f-:p)^((Rev f-:p)/^1) by A2,Th93 .= (Rev f:-p)^((Rev f-:p)/^1) by A2,Th94 .= Rotate(Rev f,p) by A4,Def2; end;

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