Copyright (c) 1990 Association of Mizar Users
environ vocabulary PARTFUN1, RELAT_1, FINSEQ_4, FUNCT_1, BOOLE, FUNCOP_1, PARTFUN2; notation TARSKI, XBOOLE_0, SUBSET_1, RELAT_1, RELSET_1, FUNCT_1, FUNCT_2, FUNCOP_1, PARTFUN1, FINSEQ_4; constructors FUNCOP_1, PARTFUN1, FINSEQ_4, MEMBERED, XBOOLE_0; clusters RELAT_1, FUNCT_1, RELSET_1, FUNCOP_1, MEMBERED, ZFMISC_1, XBOOLE_0; requirements SUBSET, BOOLE; definitions TARSKI, XBOOLE_0; theorems TARSKI, FUNCT_1, FUNCT_2, GRFUNC_1, FUNCOP_1, PARTFUN1, FINSEQ_4, RELAT_1, RELSET_1, XBOOLE_0; schemes FUNCT_2, XBOOLE_0; begin reserve x,y,X,Y for set; reserve C,D,E for non empty set; reserve SC for Subset of C; reserve SD for Subset of D; reserve SE for Subset of E; reserve c,c1,c2 for Element of C; reserve d,d1,d2 for Element of D; reserve e for Element of E; reserve f,f1,g for PartFunc of C,D; reserve t for PartFunc of D,C; reserve s for PartFunc of D,E; reserve h for PartFunc of C,E; reserve F for PartFunc of D,D; canceled 2; theorem Th3: dom f = dom g & (for c st c in dom f holds f/.c = g/.c) implies f = g proof assume A1: dom f = dom g & (for c st c in dom f holds f/.c = g/.c); now let x; assume A2: x in dom f; then reconsider y=x as Element of C; f/.y=g/.y by A1,A2; then (f qua Function).y = g/.y by A2,FINSEQ_4:def 4; hence (f qua Function).x = (g qua Function).x by A1,A2,FINSEQ_4:def 4; end; hence thesis by A1,FUNCT_1:9; end; theorem Th4: :: ograniczyc do implikacji y in rng f iff ex c st c in dom f & y = f/.c proof thus y in rng f implies ex c st c in dom f & y = f/.c proof assume y in rng f; then consider x such that A1: x in dom f & y = (f qua Function).x by FUNCT_1:def 5; reconsider x as Element of C by A1; take x; thus thesis by A1,FINSEQ_4:def 4; end; given c such that A2: c in dom f & y = f/.c; (f qua Function).c in rng f by A2,FUNCT_1:def 5; hence thesis by A2,FINSEQ_4:def 4; end; canceled; theorem Th6: h = s*f iff (for c holds c in dom h iff c in dom f & f/.c in dom s) & for c st c in dom h holds h/.c = s/.(f/.c) proof thus h = s*f implies (for c holds c in dom h iff c in dom f & f/.c in dom s) & (for c st c in dom h holds h/.c = s/.(f/.c)) proof assume A1: h = s*f; A2: now let c; thus c in dom h implies c in dom f & f/.c in dom s proof assume c in dom h; then c in dom f & (f qua Function).c in dom s by A1,FUNCT_1:21; hence thesis by FINSEQ_4:def 4; end; assume c in dom f & f/.c in dom s; then c in dom f & (f qua Function).c in dom s by FINSEQ_4:def 4; hence c in dom h by A1,FUNCT_1:21; end; hence for c holds c in dom h iff c in dom f & f/.c in dom s; let c; assume A3: c in dom h; then (h qua Function).c = (s qua Function).((f qua Function).c) by A1,FUNCT_1:22; then A4: h/.c = (s qua Function).((f qua Function).c) by A3,FINSEQ_4:def 4; A5: c in dom f & f/.c in dom s by A2,A3; then h/.c = (s qua Function).(f/.c) by A4,FINSEQ_4:def 4; hence thesis by A5,FINSEQ_4:def 4; end; assume that A6: for c holds c in dom h iff c in dom f & f/.c in dom s and A7: for c st c in dom h holds h/.c = s/.(f/.c); A8: now let x; thus x in dom h implies x in dom f & (f qua Function).x in dom s proof assume A9: x in dom h; then reconsider y=x as Element of C; y in dom f & f/.y in dom s by A6,A9; hence thesis by FINSEQ_4:def 4; end; thus x in dom f & (f qua Function).x in dom s implies x in dom h proof assume A10: x in dom f & (f qua Function).x in dom s; then reconsider y=x as Element of C; y in dom f & f/.y in dom s by A10,FINSEQ_4:def 4; hence thesis by A6; end; end; now let x; assume A11: x in dom h; then reconsider y=x as Element of C; A12: y in dom f & f/.y in dom s by A6,A11; h/.y = s/.(f/.y) by A7,A11; then (h qua Function).y = s/.(f/.y) by A11,FINSEQ_4:def 4; then (h qua Function).x = (s qua Function).(f/.y) by A12,FINSEQ_4:def 4; hence (h qua Function).x = (s qua Function).((f qua Function).x) by A12,FINSEQ_4:def 4; end; hence thesis by A8,FUNCT_1:20; end; canceled 2; theorem Th9: c in dom f & f/.c in dom s implies (s*f)/.c = s/.(f/.c) proof assume c in dom f & f/.c in dom s; then c in dom (s*f) by Th6; hence thesis by Th6; end; theorem rng f c= dom s & c in dom f implies (s*f)/.c = s/.(f/.c) proof assume A1: rng f c= dom s & c in dom f; then f/.c in rng f by Th4; hence thesis by A1,Th9; end; definition let D; let SD; redefine func id SD -> PartFunc of D,D; coherence proof dom id SD = SD & rng id SD = SD by RELAT_1:71; hence thesis by RELSET_1:11; end; end; canceled; theorem Th12: F = id SD iff dom F = SD & for d st d in SD holds F/.d = d proof thus F = id SD implies dom F = SD & for d st d in SD holds F/.d = d proof assume A1: F = id SD; hence A2: dom F = SD by RELAT_1:71; let d; assume A3: d in SD; then (F qua Function).d = d by A1,FUNCT_1:35; hence thesis by A2,A3,FINSEQ_4:def 4; end; assume A4: dom F = SD & for d st d in SD holds F/.d = d; now let x; assume A5: x in SD; then reconsider x1=x as Element of D; F/.x1=x1 by A4,A5; hence (F qua Function).x = x by A4,A5,FINSEQ_4:def 4; end; hence thesis by A4,FUNCT_1:34; end; canceled; theorem d in dom F /\ SD implies F/.d = (F*id SD)/.d proof assume A1: d in dom F /\ SD; then A2: d in dom F & d in SD by XBOOLE_0:def 3; then A3: (id SD)/.d in dom F by Th12; d in dom id SD by A2,RELAT_1:71; then A4: d in dom (F*(id SD)) by A3,Th6; (F qua Function).d = ((F*(id SD)) qua Function).d by A1,FUNCT_1:38; then F/.d = ((F*(id SD)) qua Function).d by A2,FINSEQ_4:def 4; hence thesis by A4,FINSEQ_4:def 4; end; theorem d in dom((id SD)*F) iff d in dom F & F/.d in SD proof thus d in dom((id SD)*F) implies d in dom F & F/.d in SD proof assume d in dom((id SD)*F); then d in dom F & F/.d in dom (id SD) by Th6; hence thesis by RELAT_1:71; end; assume d in dom F & F/.d in SD; then d in dom F & F/.d in dom (id SD) by RELAT_1:71; hence thesis by Th6; end; theorem (for c1,c2 st c1 in dom f & c2 in dom f & f/.c1 = f/.c2 holds c1 = c2) implies f is one-to-one proof assume A1: for c1,c2 st c1 in dom f & c2 in dom f & f/.c1 = f/.c2 holds c1 = c2; now let x,y; assume A2: x in dom f & y in dom f & (f qua Function).x = (f qua Function).y; then reconsider x1 = x as Element of C; reconsider y1 = y as Element of C by A2; x1 in dom f & y1 in dom f & f/.x1 = (f qua Function).y1 by A2,FINSEQ_4:def 4; then x1 in dom f & y1 in dom f & f/.x1 = f/.y1 by FINSEQ_4:def 4; hence x=y by A1; end; hence thesis by FUNCT_1:def 8; end; theorem f is one-to-one & x in dom f & y in dom f & f/.x = f/.y implies x = y proof assume that A1: f is one-to-one and A2: x in dom f and A3: y in dom f; assume f/.x = f/.y; then f.x = f/.y by A2,FINSEQ_4:def 4 .= f.y by A3,FINSEQ_4:def 4; hence x = y by A1,A2,A3,FUNCT_1:def 8; end; definition cluster {} -> one-to-one; coherence; end; definition let X,Y; cluster one-to-one PartFunc of X,Y; existence proof {} is PartFunc of X,Y by PARTFUN1:56; hence thesis; end; end; definition let X,Y; let f be one-to-one PartFunc of X,Y; redefine func f" -> PartFunc of Y,X; coherence by PARTFUN1:39; end; theorem Th18: for f being one-to-one PartFunc of C,D holds for g be PartFunc of D,C holds g = f" iff dom g = rng f & for d,c holds d in rng f & c = g/.d iff c in dom f & d = f/.c proof let f be one-to-one PartFunc of C,D; let g be PartFunc of D,C; thus g = f" implies dom g = rng f & for d,c holds d in rng f & c = g/.d iff c in dom f & d = f/.c proof assume A1: g = f"; then A2: dom g = rng f & for y,x holds y in rng f & x = (g qua Function).y iff x in dom f & y = (f qua Function).x by FUNCT_1:54; thus dom g = rng f by A1,FUNCT_1:54; let d,c; thus d in rng f & c = g/.d implies c in dom f & d = f/.c proof assume A3: d in rng f & c = g/.d; then c = (g qua Function).d by A2,FINSEQ_4:def 4; then c in dom f & d = (f qua Function).c by A1,A3,FUNCT_1:54; hence thesis by FINSEQ_4:def 4; end; assume c in dom f & d = f/.c; then c in dom f & d = (f qua Function).c by FINSEQ_4:def 4; then d in rng f & c = (g qua Function).d by A1,FUNCT_1:54; hence thesis by A2,FINSEQ_4:def 4; end; assume that A4: dom g = rng f and A5: for d,c holds d in rng f & c = g/.d iff c in dom f & d = f/.c and A6: g <> f"; now per cases by A6,Th3; suppose dom (f") <> dom g; hence contradiction by A4,FUNCT_1:55; suppose ex d st d in dom (f") & f"/.d <> g/.d; then consider d such that A7: d in dom (f") & f"/.d <> g/.d; A8: d in rng f by A7,FUNCT_1:55; f"/.d in rng (f") by A7,Th4; then A9: f"/.d in dom f by FUNCT_1:55; d = (f qua Function).(((f") qua Function).d) by A8,FUNCT_1:57; then d = (f qua Function).(f"/.d) by A7,FINSEQ_4:def 4; then d = f/.(f"/.d) by A9,FINSEQ_4:def 4; hence contradiction by A5,A7,A9; end; hence contradiction; end; canceled 3; theorem for f being one-to-one PartFunc of C,D st c in dom f holds c = f"/.(f/.c) & c = (f"*f)/.c proof let f be one-to-one PartFunc of C,D; assume A1: c in dom f; f" = f"; then A2: f/.c in rng f & c = f"/.(f/.c) by A1,Th18; thus A3: c = f"/.(f/.c) by A1,Th18; f/.c in dom (f") & c in dom f by A1,A2,FUNCT_1:55; hence thesis by A3,Th9; end; theorem for f being one-to-one PartFunc of C,D st d in rng f holds d = f/.(f"/.d) & d = (f*(f"))/.d proof let f be one-to-one PartFunc of C,D; assume A1: d in rng f; then A2: d = (f qua Function).(((f") qua Function).d) & d = ((f*f") qua Function).d by FUNCT_1:57; d in dom (f*f") by A1,FUNCT_1:59; hence thesis by A1,A2,Th18,FINSEQ_4:def 4; end; theorem f is one-to-one & dom f = rng t & rng f = dom t & (for c,d st c in dom f & d in dom t holds f/.c = d iff t/.d = c) implies t = f" proof assume A1: f is one-to-one & dom f = rng t & rng f = dom t & (for c,d st c in dom f & d in dom t holds f/.c = d iff t/.d = c); now let x,y; assume A2: x in dom f & y in dom t; then reconsider x1=x as Element of C; reconsider y1=y as Element of D by A2; thus (f qua Function).x = y implies (t qua Function).y = x proof assume (f qua Function).x = y; then f/.x1 = y1 by A2,FINSEQ_4:def 4; then t/.y1 = x1 by A1,A2; hence thesis by A2,FINSEQ_4:def 4; end; assume (t qua Function).y = x; then t/.y1 = x1 by A2,FINSEQ_4:def 4; then f/.x1 = y1 by A1,A2; hence (f qua Function).x = y by A2,FINSEQ_4:def 4; end; hence thesis by A1,FUNCT_1:60; end; canceled 7; theorem Th32: g = f|X iff dom g = dom f /\ X & for c st c in dom g holds g/.c = f/.c proof thus g = f|X implies dom g = dom f /\ X & for c st c in dom g holds g/.c = f/.c proof assume A1: g = f|X; then A2: dom g = dom f /\ X & for x st x in dom g holds (g qua Function).x = (f qua Function).x by FUNCT_1:68; thus dom g = dom f /\ X by A1,FUNCT_1:68; let c; assume A3: c in dom g; then A4: c in dom f by A2,XBOOLE_0:def 3; (g qua Function).c = (f qua Function).c by A1,A3,FUNCT_1:68; then g/.c = (f qua Function).c by A3,FINSEQ_4:def 4; hence thesis by A4,FINSEQ_4:def 4; end; assume A5: dom g = dom f /\ X & for c st c in dom g holds g/.c = f/.c; now let x; assume A6: x in dom g; then A7: x in dom f by A5,XBOOLE_0:def 3; reconsider y=x as Element of C by A6; g/.y = f/.y by A5,A6; then (g qua Function).y = f/.y by A6,FINSEQ_4:def 4; hence (g qua Function).x = (f qua Function).x by A7,FINSEQ_4:def 4; end; hence thesis by A5,FUNCT_1:68; end; canceled; theorem Th34: c in dom f /\ X implies f|X/.c = f/.c proof assume c in dom f /\ X; then c in dom (f|X) by RELAT_1:90; hence thesis by Th32; end; theorem c in dom f & c in X implies f|X/.c = f/.c proof assume c in dom f & c in X; then c in dom f /\ X by XBOOLE_0:def 3; hence thesis by Th34; end; theorem c in dom f & c in X implies f/.c in rng (f|X) proof assume A1: c in dom f & c in X; then (f qua Function).c in rng (f|X) by FUNCT_1:73; hence thesis by A1,FINSEQ_4:def 4; end; definition let C,D; let X,f; redefine func X|f -> PartFunc of C,D; coherence by PARTFUN1:46; end; theorem Th37: g = X|f iff (for c holds c in dom g iff c in dom f & f/.c in X) & (for c st c in dom g holds g/.c = f/.c) proof thus g = X|f implies (for c holds c in dom g iff c in dom f & f/.c in X) & (for c st c in dom g holds g/.c = f/.c) proof assume A1: g = X|f; now let c; thus c in dom g implies c in dom f & f/.c in X proof assume c in dom g; then c in dom f & (f qua Function).c in X by A1,FUNCT_1:86; hence thesis by FINSEQ_4:def 4; end; assume c in dom f & f/.c in X; then c in dom f & (f qua Function).c in X by FINSEQ_4:def 4; hence c in dom g by A1,FUNCT_1:86; end; hence for c holds c in dom g iff c in dom f & f/.c in X; let c; assume A2: c in dom g; then (g qua Function).c = (f qua Function).c by A1,FUNCT_1:87; then A3: g/.c = (f qua Function).c by A2,FINSEQ_4:def 4; c in dom f by A1,A2,FUNCT_1:86; hence thesis by A3,FINSEQ_4:def 4; end; assume A4: (for c holds c in dom g iff c in dom f & f/.c in X) & for c st c in dom g holds g/.c = f/.c; A5:now let x; thus x in dom g implies x in dom f & (f qua Function).x in X proof assume A6: x in dom g; then reconsider x1=x as Element of C; x1 in dom f & f/.x1 in X by A4,A6; hence thesis by FINSEQ_4:def 4; end; assume A7: x in dom f & (f qua Function).x in X; then reconsider x1=x as Element of C; x1 in dom f & f/.x1 in X by A7,FINSEQ_4:def 4; hence x in dom g by A4; end; now let x; assume A8: x in dom g; then reconsider x1=x as Element of C; A9: x1 in dom f by A4,A8; g/.x1 = f/.x1 by A4,A8; then (g qua Function).x1 = f/.x1 by A8,FINSEQ_4:def 4; hence (g qua Function).x = (f qua Function).x by A9,FINSEQ_4:def 4; end; hence thesis by A5,FUNCT_1:85; end; theorem c in dom (X|f) iff c in dom f & f/.c in X by Th37; theorem c in dom (X|f) implies X|f/.c = f/.c by Th37; theorem Th40: SD = f.: X iff for d holds d in SD iff ex c st c in dom f & c in X & d = f/.c proof thus SD = f.:X implies for d holds d in SD iff ex c st c in dom f & c in X & d = f/.c proof assume A1: SD = f.:X; let d; thus d in SD implies ex c st c in dom f & c in X & d = f/.c proof assume d in SD; then consider x such that A2: x in dom f & x in X & d = (f qua Function).x by A1,FUNCT_1:def 12; reconsider x as Element of C by A2; take x; thus x in dom f & x in X by A2; thus thesis by A2,FINSEQ_4:def 4; end; given c such that A3: c in dom f & c in X & d = f/.c; f/.c = (f qua Function).c by A3,FINSEQ_4:def 4; hence thesis by A1,A3,FUNCT_1:def 12; end; assume that A4: for d holds d in SD iff ex c st c in dom f & c in X & d = f/.c and A5: SD <> f.:X; consider x such that A6: not (x in SD iff x in f.:X) by A5,TARSKI:2; now per cases by A6; suppose A7: x in SD & not x in f.:X; then reconsider x as Element of D; consider c such that A8: c in dom f & c in X & x = f/.c by A4,A7; x = (f qua Function).c by A8,FINSEQ_4:def 4; hence contradiction by A7,A8,FUNCT_1:def 12; suppose A9: x in f.:X & not x in SD; then consider y such that A10: y in dom f & y in X & x = (f qua Function).y by FUNCT_1:def 12; reconsider y as Element of C by A10; x = f/.y by A10,FINSEQ_4:def 4; hence contradiction by A4,A9,A10; end; hence contradiction; end; theorem d in (f qua Relation of C,D).:X iff ex c st c in dom f & c in X & d = f/.c by Th40; theorem c in dom f implies f.:{c} = {f/.c} proof assume A1: c in dom f; hence f.:{c} = {(f qua Function).c} by FUNCT_1:117 .= {f/.c} by A1,FINSEQ_4:def 4; end; theorem c1 in dom f & c2 in dom f implies f.:{c1,c2} = {f/.c1,f/.c2} proof assume A1: c1 in dom f & c2 in dom f; hence f.:{c1,c2} = {(f qua Function).c1,(f qua Function).c2} by FUNCT_1:118 .= {f/.c1,(f qua Function).c2} by A1,FINSEQ_4:def 4 .= {f/.c1,f/.c2} by A1,FINSEQ_4:def 4; end; theorem Th44: SC = f"X iff for c holds c in SC iff c in dom f & f/.c in X proof thus SC = f"X implies for c holds c in SC iff c in dom f & f/.c in X proof assume A1: SC = f"X; let c; thus c in SC implies c in dom f & f/.c in X proof assume c in SC; then c in dom f & (f qua Function).c in X by A1,FUNCT_1:def 13; hence thesis by FINSEQ_4:def 4; end; assume c in dom f & f/.c in X; then c in dom f & (f qua Function).c in X by FINSEQ_4:def 4; hence c in SC by A1,FUNCT_1:def 13; end; assume A2: for c holds c in SC iff c in dom f & f/.c in X; now let x; thus x in SC implies x in dom f & (f qua Function).x in X proof assume A3: x in SC; then reconsider x1=x as Element of C; x1 in dom f & f/.x1 in X by A2,A3; hence thesis by FINSEQ_4:def 4; end; assume A4: x in dom f & (f qua Function).x in X; then reconsider x1=x as Element of C; x1 in dom f & f/.x1 in X by A4,FINSEQ_4:def 4; hence x in SC by A2; end; hence thesis by FUNCT_1:def 13; end; canceled; theorem for f ex g being Function of C,D st for c st c in dom f holds g.c = f/.c proof let f; consider g being Function of C,D such that A1: for x st x in dom f holds g.x = (f qua Function).x by FUNCT_2:136; take g; let c; assume A2: c in dom f; then g.c = (f qua Function).c by A1; hence thesis by A2,FINSEQ_4:def 4; end; theorem f tolerates g iff for c st c in dom f /\ dom g holds f/.c = g/.c proof thus f tolerates g implies for c st c in dom f /\ dom g holds f/.c = g/.c proof assume A1: f tolerates g; let c; assume A2: c in dom f /\ dom g; then A3: c in dom f & c in dom g by XBOOLE_0:def 3; (f qua Function).c = (g qua Function).c by A1,A2,PARTFUN1:def 6; then f/.c = (g qua Function).c by A3,FINSEQ_4:def 4; hence thesis by A3,FINSEQ_4:def 4; end; assume A4: for c st c in dom f /\ dom g holds f/.c = g/.c; now let x; assume A5: x in dom f /\ dom g; then A6: x in dom f & x in dom g by XBOOLE_0:def 3; reconsider x1=x as Element of C by A5; f/.x1 = g/.x1 by A4,A5; then (f qua Function).x = g/.x1 by A6,FINSEQ_4:def 4; hence (f qua Function).x = (g qua Function).x by A6,FINSEQ_4:def 4; end; hence thesis by PARTFUN1:def 6; end; scheme PartFuncExD{D,C()->non empty set, P[set,set]}: ex f being PartFunc of D(),C() st (for d be Element of D() holds d in dom f iff (ex c be Element of C() st P[d,c])) & (for d be Element of D() st d in dom f holds P[d,f/.d]) proof consider x being Element of C(); defpred Q[set,set] means ((ex c be Element of C() st P[$1,c]) implies P[$1,$2]) & ((for c be Element of C() holds not P[$1,c]) implies $2=x); A1: for d be Element of D() ex z be Element of C() st Q[d,z] proof let d be Element of D(); (for c be Element of C() holds not P[d,c]) implies ex z be Element of C() st ((ex c be Element of C() st P[d,c]) implies P[d,z]) & ((for c be Element of C() holds not P[d,c]) implies z=x); hence thesis; end; consider g being Function of D(),C() such that A2: for d be Element of D() holds Q[d,g.d] from FuncExD(A1); A3: dom g = D() & rng g c= C() by FUNCT_2:def 1,RELSET_1:12; defpred R[set] means ex c be Element of C() st P[$1,c]; consider X be set such that A4: for x holds x in X iff x in D() & R[x] from Separation; for x holds x in X implies x in D() by A4; then reconsider X as Subset of D() by TARSKI:def 3; reconsider f=g|X as PartFunc of D(),C(); take f; thus for d be Element of D() holds d in dom f iff (ex c be Element of C() st P[d,c]) proof let d be Element of D(); dom f c= X by RELAT_1:87; hence d in dom f implies ex c be Element of C() st P[d,c] by A4; assume ex c be Element of C() st P[d,c]; then d in X & d in D() by A4; then d in dom g /\ X by A3,XBOOLE_0:def 3; hence thesis by RELAT_1:90; end; let d be Element of D(); assume A5: d in dom f; dom f c= X by RELAT_1:87; then ex c be Element of C() st P[d,c] by A4,A5; then P[d,g.d] by A2; then P[d,(f qua Function).d] by A5,FUNCT_1:70; hence thesis by A5,FINSEQ_4:def 4; end; scheme LambdaPFD{D,C()->non empty set, F(set)->Element of C(), P[set]}: ex f being PartFunc of D(),C() st (for d be Element of D() holds d in dom f iff P[d]) & (for d be Element of D() st d in dom f holds f/.d = F(d)) proof defpred Q[set,set] means P[$1] & $2 = F($1); consider f being PartFunc of D(),C() such that A1: for d be Element of D() holds d in dom f iff ex c be Element of C() st Q[d,c] and A2: for d be Element of D() st d in dom f holds Q[d,f/.d] from PartFuncExD; take f; thus for d be Element of D() holds d in dom f iff P[d] proof let d be Element of D(); thus d in dom f implies P[d] proof assume d in dom f; then ex c be Element of C() st P[d] & c = F(d) by A1; hence thesis; end; assume P[d]; then ex c be Element of C() st P[d] & c = F(d); hence thesis by A1; end; thus thesis by A2; end; scheme UnPartFuncD{C,D()->non empty set, X()->set, F(set)->Element of D()}: for f,g being PartFunc of C(),D() st (dom f=X() & for c be Element of C() st c in dom f holds f/.c = F(c)) & (dom g=X() & for c be Element of C() st c in dom g holds g/.c = F(c)) holds f = g proof let f,g be PartFunc of C(),D(); assume that A1: (dom f=X() & for c be Element of C() st c in dom f holds f/.c = F(c)) and A2: dom g=X() & for c be Element of C() st c in dom g holds g/.c = F(c); now let c be Element of C(); assume A3: c in dom f; hence f/.c = F(c) by A1 .= g/.c by A1,A2,A3; end; hence thesis by A1,A2,Th3; end; definition let C,D; let SC,d; redefine func SC --> d -> PartFunc of C,D; coherence proof now per cases; suppose SC = {}; then dom (SC --> d) = {} by FUNCOP_1:16; hence thesis by PARTFUN1:55; suppose SC <> {}; then dom (SC --> d) = SC & rng (SC --> d) = {d} by FUNCOP_1:14,19; then (SC --> d) is PartFunc of SC,{d} by PARTFUN1:24; then A1: (SC --> d) is PartFunc of C,{d} by PARTFUN1:30; now let x; assume x in {d}; then x = d by TARSKI:def 1; hence x in D; end; then {d} c= D by TARSKI:def 3; hence thesis by A1,PARTFUN1:31; end; hence thesis; end; end; theorem Th48: c in SC implies (SC --> d)/.c = d proof assume A1: c in SC; A2: dom (SC --> d) = SC by FUNCOP_1:19; ((SC --> d) qua Function).c = d by A1,FUNCOP_1:13; hence thesis by A1,A2,FINSEQ_4:def 4; end; theorem (for c st c in dom f holds f/.c = d) implies f = dom f --> d proof assume A1: (for c st c in dom f holds f/.c = d); now let x; assume A2: x in dom f; then reconsider x1=x as Element of C; f/.x1 = d by A1,A2; hence ( f qua Function).x = d by A2,FINSEQ_4:def 4; end; hence thesis by FUNCOP_1:17; end; theorem c in dom f implies f*(SE --> c) = SE --> f/.c proof assume A1: c in dom f; then f*(SE --> c) = SE --> (f qua Function).c by FUNCOP_1:23; hence thesis by A1,FINSEQ_4:def 4; end; theorem (id SC) is total iff SC = C proof thus (id SC) is total implies SC = C proof assume (id SC) is total; then dom (id SC) = C by PARTFUN1:def 4; hence SC = C by RELAT_1:71; end; assume SC = C; then dom (id SC) = C by RELAT_1:71; hence (id SC) is total by PARTFUN1:def 4; end; theorem (SC --> d) is total implies SC <> {} proof assume that A1: (SC --> d) is total and A2: SC = {}; dom (SC --> d) = C by A1,PARTFUN1:def 4; hence contradiction by A2,FUNCOP_1:16; end; theorem (SC --> d) is total iff SC = C proof thus (SC --> d) is total implies SC = C proof assume (SC --> d) is total; then dom (SC --> d) = C by PARTFUN1:def 4; hence SC = C by FUNCOP_1:19; end; assume SC = C; then dom (SC --> d) = C by FUNCOP_1:19; hence (SC --> d) is total by PARTFUN1:def 4; end; :: :: PARTIAL FUNCTION CONSTANT ON SET :: definition let C,D; let f,X; canceled 2; pred f is_constant_on X means :Def3: ex d st for c st c in X /\ dom f holds f/.c = d; end; canceled; theorem f is_constant_on X iff for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f holds f/.c1=f/.c2 proof thus f is_constant_on X implies for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f holds f/.c1=f/.c2 proof assume f is_constant_on X; then consider d such that A1: for c st c in X /\ dom f holds f/.c = d by Def3; let c1,c2; assume c1 in X /\ dom f & c2 in X /\ dom f; then f/.c1 = d & f/.c2 = d by A1; hence thesis; end; assume A2: for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f holds f/.c1=f/.c2; now per cases; suppose A3: X /\ dom f = {}; now consider d being Element of D; take d; let c; thus c in X /\ dom f implies f/.c = d by A3; end; hence thesis by Def3; suppose A4: X /\ dom f <> {}; consider x being Element of X /\ dom f; x in dom f by A4,XBOOLE_0:def 3; then reconsider x as Element of C; for c holds c in X /\ dom f implies f/.c = f/.x by A2; hence thesis by Def3; end; hence thesis; end; theorem X meets dom f implies (f is_constant_on X iff ex d st rng (f|X) = {d}) proof assume A1: X /\ dom f <> {}; thus f is_constant_on X implies ex d st rng (f|X) = {d} proof assume f is_constant_on X; then consider d such that A2: for c st c in X /\ dom f holds f/.c = d by Def3; take d; thus rng (f|X) c= {d} proof let x; assume x in rng (f|X); then consider y such that A3: y in dom (f|X) & ((f|X) qua Function).y = x by FUNCT_1:def 5; reconsider y as Element of C by A3; dom (f|X) = X /\ dom f by RELAT_1:90; then d = f/.y by A2,A3 .= f|X/.y by A3,Th32 .= x by A3,FINSEQ_4:def 4; hence x in {d} by TARSKI:def 1; end; thus {d} c= rng (f|X) proof let x such that A4: x in {d}; consider y being Element of X /\ dom f; y in dom f by A1,XBOOLE_0:def 3; then reconsider y as Element of C; A5: dom (f|X) = X /\ dom f by RELAT_1:90; then f|X/.y = f/.y by A1,Th32 .= d by A1,A2 .= x by A4,TARSKI:def 1; hence x in rng (f|X) by A1,A5,Th4; end; end; given d such that A6: rng (f|X) = {d}; take d; let c; assume c in X /\ dom f; then A7: c in dom (f|X) by RELAT_1:90; then f|X/.c in {d} by A6,Th4; then f|X/.c = d by TARSKI:def 1; hence thesis by A7,Th32; end; theorem f is_constant_on X & Y c= X implies f is_constant_on Y proof assume A1: f is_constant_on X & Y c= X; then consider d such that A2: for c st c in X /\ dom f holds f/.c = d by Def3; now let c; assume c in Y /\ dom f; then c in Y & c in dom f by XBOOLE_0:def 3; then c in X /\ dom f by A1,XBOOLE_0:def 3; hence f/.c = d by A2; end; hence thesis by Def3; end; theorem Th58: X misses dom f implies f is_constant_on X proof assume A1: X /\ dom f = {}; now consider d being Element of D; take d; let c; thus c in X /\ dom f implies f/.c = d by A1; end; hence thesis by Def3; end; theorem f|SC = dom (f|SC) --> d implies f is_constant_on SC proof assume A1: f|SC = dom (f|SC) --> d; now let c; assume c in SC /\ dom f; then A2: c in dom (f|SC) by RELAT_1:90; then f|SC/.c = d by A1,Th48; hence f/.c = d by A2,Th32; end; hence f is_constant_on SC by Def3; end; theorem f is_constant_on {x} proof now per cases; suppose {x} /\ dom f = {}; then {x} misses dom f by XBOOLE_0:def 7; hence f is_constant_on {x} by Th58; suppose A1: {x} /\ dom f <> {}; consider y being Element of {x} /\ dom f; y in {x} & y in dom f by A1,XBOOLE_0:def 3; then reconsider x1=x as Element of C by TARSKI:def 1; now take d = f/.x1; let c; assume c in {x} /\ dom f; then c in {x} by XBOOLE_0:def 3; hence f/.c = f/.x1 by TARSKI:def 1; end; hence f is_constant_on {x} by Def3; end; hence thesis; end; theorem f is_constant_on X & f is_constant_on Y & X /\ Y meets dom f implies f is_constant_on X \/ Y proof assume A1: f is_constant_on X & f is_constant_on Y & X /\ Y /\ dom f <> {}; then consider d1 such that A2: for c st c in X /\ dom f holds f/.c = d1 by Def3; consider d2 such that A3: for c st c in Y /\ dom f holds f/.c = d2 by A1,Def3; consider x being Element of X /\ Y /\ dom f; A4: x in X /\ Y & x in dom f by A1,XBOOLE_0:def 3; then reconsider x as Element of C; take d1; x in X & x in dom f by A4,XBOOLE_0:def 3; then x in X /\ dom f by XBOOLE_0:def 3; then A5: f/.x = d1 by A2; x in Y & x in dom f by A4,XBOOLE_0:def 3; then x in Y /\ dom f by XBOOLE_0:def 3; then A6: d1 = d2 by A3,A5; let c; assume c in (X \/ Y) /\ dom f; then A7: c in X \/ Y & c in dom f by XBOOLE_0:def 3; now per cases by A7,XBOOLE_0:def 2; suppose c in X; then c in X /\ dom f by A7,XBOOLE_0:def 3; hence f/.c = d1 by A2; suppose c in Y; then c in Y /\ dom f by A7,XBOOLE_0:def 3; hence f/.c = d1 by A3,A6; end; hence f/.c = d1; end; theorem f is_constant_on Y implies f|X is_constant_on Y proof assume f is_constant_on Y; then consider d such that A1: for c st c in Y /\ dom f holds f/.c = d by Def3; take d; let c; assume c in Y /\ dom (f|X); then A2: c in Y & c in dom (f|X) by XBOOLE_0:def 3; then c in Y & c in dom f /\ X by RELAT_1:90; then c in Y & c in dom f by XBOOLE_0:def 3; then c in Y /\ dom f by XBOOLE_0:def 3; then f/.c = d by A1; hence f|X/.c = d by A2,Th32; end; theorem SC --> d is_constant_on SC proof take d; let c; assume c in SC /\ dom (SC --> d); then c in SC by XBOOLE_0:def 3; hence (SC --> d)/.c = d by Th48; end; :: :: OF PARTIAL FUNCTION FROM A DOMAIN TO A DOMAIN :: theorem f c= g iff dom f c= dom g & (for c st c in dom f holds f/.c = g/.c) proof thus f c= g implies dom f c= dom g & (for c st c in dom f holds f/.c = g/.c) proof assume A1: f c= g; then A2: (dom f c= dom g) & for x st x in dom f holds (f qua Function).x = (g qua Function).x by GRFUNC_1:8; thus dom f c= dom g by A1,GRFUNC_1:8; let c; assume A3: c in dom f; then (f qua Function).c = (g qua Function).c by A1,GRFUNC_1:8; then f/.c = (g qua Function).c by A3,FINSEQ_4:def 4; hence thesis by A2,A3,FINSEQ_4:def 4; end; assume A4: dom f c= dom g & for c st c in dom f holds f/.c = g/.c; now let x; assume A5: x in dom f; then reconsider x1=x as Element of C; f/.x1 = g/.x1 by A4,A5; then (f qua Function).x = g/.x1 by A5,FINSEQ_4:def 4; hence (f qua Function).x = (g qua Function).x by A4,A5,FINSEQ_4:def 4; end; hence thesis by A4,GRFUNC_1:8; end; theorem Th65: c in dom f & d = f/.c iff [c,d] in f proof thus c in dom f & d = f/.c implies [c,d] in f proof assume c in dom f & d = f/.c; then c in dom f & d = (f qua Function).c by FINSEQ_4:def 4; hence thesis by FUNCT_1:8; end; assume [c,d] in f; then c in dom f & d = (f qua Function).c by FUNCT_1:8; hence thesis by FINSEQ_4:def 4; end; theorem [c,e] in (s*f) implies [c,f/.c] in f & [f/.c,e] in s proof assume [c,e] in (s*f); then A1: [c,(f qua Function).c] in f & [(f qua Function).c,e] in s by GRFUNC_1:12; then c in dom f by FUNCT_1:8; hence thesis by A1,FINSEQ_4:def 4; end; theorem f = {[c,d]} implies f/.c = d proof assume A1: f = {[c,d]}; then [c,d] in f by TARSKI:def 1; then A2:c in dom f by FUNCT_1:8; (f qua Function).c = d by A1,GRFUNC_1:16; hence thesis by A2,FINSEQ_4:def 4; end; theorem dom f = {c} implies f = {[c,f/.c]} proof assume A1: dom f = {c}; then A2: c in dom f by TARSKI:def 1; f = {[c,(f qua Function).c]} by A1,GRFUNC_1:18; hence thesis by A2,FINSEQ_4:def 4; end; theorem f1 = f /\ g & c in dom f1 implies f1/.c = f/.c & f1/.c = g/.c proof assume A1: f1 = f /\ g & c in dom f1; then [c,(f1 qua Function).c] in f1 by FUNCT_1:8; then [c,(f1 qua Function).c] in f & [c,(f1 qua Function).c] in g by A1,XBOOLE_0:def 3; then A2: c in dom f & c in dom g by FUNCT_1:8; (f1 qua Function).c = (f qua Function).c & (f1 qua Function).c = (g qua Function).c by A1,GRFUNC_1:29; then f1/.c = (f qua Function).c & f1/.c = (g qua Function).c by A1,FINSEQ_4:def 4; hence thesis by A2,FINSEQ_4:def 4; end; theorem c in dom f & f1 = f \/ g implies f1/.c = f/.c proof assume A1: c in dom f & f1 = f \/ g; then [c,(f qua Function).c] in f by FUNCT_1:8; then [c,(f qua Function).c] in f1 by A1,XBOOLE_0:def 2; then A2: c in dom f1 by FUNCT_1:8; (f1 qua Function).c = (f qua Function).c by A1,GRFUNC_1:34; then f1/.c = (f qua Function).c by A2,FINSEQ_4:def 4; hence thesis by A1,FINSEQ_4:def 4; end; theorem c in dom g & f1 = f \/ g implies f1/.c = g/.c proof assume A1: c in dom g & f1 = f \/ g; then [c,(g qua Function).c] in g by FUNCT_1:8; then [c,(g qua Function).c] in f1 by A1,XBOOLE_0:def 2; then A2: c in dom f1 by FUNCT_1:8; (f1 qua Function).c = (g qua Function).c by A1,GRFUNC_1:35; then f1/.c = (g qua Function).c by A2,FINSEQ_4:def 4; hence thesis by A1,FINSEQ_4:def 4; end; theorem c in dom f1 & f1 = f \/ g implies f1/.c = f/.c or f1/.c = g/.c proof assume A1: c in dom f1 & f1 = f \/ g; then [c,f1/.c] in f1 by Th65; then A2: [c,f1/.c] in f or [c,f1/.c] in g by A1,XBOOLE_0:def 2; now per cases by A2,FUNCT_1:8; suppose c in dom f; then [c,f/.c] in f by Th65; then [c,f/.c] in f1 by A1,XBOOLE_0:def 2; hence thesis by Th65; suppose c in dom g; then [c,g/.c] in g by Th65; then [c,g/.c] in f1 by A1,XBOOLE_0:def 2; hence thesis by Th65; end; hence thesis; end; theorem c in dom f & c in SC iff [c,f/.c] in (f|SC) proof thus c in dom f & c in SC implies [c,f/.c] in (f|SC) proof assume A1: c in dom f & c in SC; then [c,(f qua Function).c] in (f|SC) by GRFUNC_1:52; hence thesis by A1,FINSEQ_4:def 4; end; assume [c,f/.c] in (f|SC); then c in dom (f|SC) by FUNCT_1:8; then c in dom f /\ SC by RELAT_1:90; hence thesis by XBOOLE_0:def 3; end; theorem c in dom f & f/.c in SD iff [c,f/.c] in (SD|f) proof thus c in dom f & f/.c in SD implies [c,f/.c] in (SD|f) proof assume A1: c in dom f & f/.c in SD; then c in dom f & (f qua Function).c in SD by FINSEQ_4:def 4; then [c,(f qua Function).c] in (SD|f) by GRFUNC_1:67; hence thesis by A1,FINSEQ_4:def 4; end; assume [c,f/.c] in (SD|f); then c in dom (SD|f) by FUNCT_1:8; then c in dom f & (f qua Function).c in SD by FUNCT_1:86; hence thesis by FINSEQ_4:def 4; end; theorem c in f"SD iff [c,f/.c] in f & f/.c in SD proof thus c in f"SD implies [c,f/.c] in f & f/.c in SD proof assume c in f"SD; then A1: [c,(f qua Function).c] in f & (f qua Function).c in SD by GRFUNC_1:87; then c in dom f by FUNCT_1:8; hence thesis by A1,FINSEQ_4:def 4; end; assume A2: [c,f/.c] in f & f/.c in SD; then c in dom f by Th65; hence thesis by A2,Th44; end; theorem Th76: f is_constant_on X iff ex d st for c st c in X /\ dom f holds f.c = d proof hereby assume f is_constant_on X; then consider d such that A1: for c st c in X /\ dom f holds f/.c = d by Def3; take d; let c; assume A2: c in X /\ dom f; then c in dom f by XBOOLE_0:def 3; hence f.c = f/.c by FINSEQ_4:def 4 .= d by A1,A2; end; given d such that A3: for c st c in X /\ dom f holds f.c = d; take d; let c; assume A4: c in X /\ dom f; then c in dom f by XBOOLE_0:def 3; hence f/.c = f.c by FINSEQ_4:def 4 .= d by A3,A4; end; theorem f is_constant_on X iff for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f holds f.c1=f.c2 proof thus f is_constant_on X implies for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f holds f.c1=f.c2 proof assume f is_constant_on X; then consider d such that A1: for c st c in X /\ dom f holds f.c = d by Th76; let c1,c2; assume c1 in X /\ dom f & c2 in X /\ dom f; then f.c1 = d & f.c2 = d by A1; hence thesis; end; assume A2: for c1,c2 st c1 in X /\ dom f & c2 in X /\ dom f holds f.c1=f.c2; now per cases; suppose A3: X /\ dom f = {}; now consider d being Element of D; take d; let c; thus c in X /\ dom f implies f.c = d by A3; end; hence thesis by Th76; suppose A4: X /\ dom f <> {}; consider x being Element of X /\ dom f; now let c; assume A5: c in X /\ dom f; A6: x in dom f by A4,XBOOLE_0:def 3; hence f.c = f.x by A2,A5 .= f/.x by A6,FINSEQ_4:def 4; end; hence thesis by Th76; end; hence thesis; end; theorem d in f.:X implies ex c st c in dom f & c in X & d = f.c proof assume d in f.:X; then consider x such that A1: x in dom f & x in X & d = (f qua Function).x by FUNCT_1:def 12; reconsider x as Element of C by A1; take x; thus thesis by A1; end; theorem f is one-to-one implies (d in rng f & c = (f").d iff c in dom f & d = f.c) proof assume A1: f is one-to-one; f" = (f qua Function)"; hence thesis by A1,FUNCT_1:54; end;