let T1, T2 be DecoratedTree of NAT ; :: thesis: ( T1 . {} = FirstLoc M & ( for t being Element of dom T1 holds
( succ t = { (t ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T1 . t)),(T1 . t))) } & ( for m being Element of NAT st m in card (NIC ((M /. (T1 . t)),(T1 . t))) holds
T1 . (t ^ <*m*>) = (LocSeq ((NIC ((M /. (T1 . t)),(T1 . t))),S)) . m ) ) ) & T2 . {} = FirstLoc M & ( for t being Element of dom T2 holds
( succ t = { (t ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T2 . t)),(T2 . t))) } & ( for m being Element of NAT st m in card (NIC ((M /. (T2 . t)),(T2 . t))) holds
T2 . (t ^ <*m*>) = (LocSeq ((NIC ((M /. (T2 . t)),(T2 . t))),S)) . m ) ) ) implies T1 = T2 )

assume that
A13: T1 . {} = FirstLoc M and
A14: for t being Element of dom T1 holds
( succ t = { (t ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T1 . t)),(T1 . t))) } & ( for m being Element of NAT st m in card (NIC ((M /. (T1 . t)),(T1 . t))) holds
T1 . (t ^ <*m*>) = (LocSeq ((NIC ((M /. (T1 . t)),(T1 . t))),S)) . m ) ) and
A15: T2 . {} = FirstLoc M and
A16: for t being Element of dom T2 holds
( succ t = { (t ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T2 . t)),(T2 . t))) } & ( for m being Element of NAT st m in card (NIC ((M /. (T2 . t)),(T2 . t))) holds
T2 . (t ^ <*m*>) = (LocSeq ((NIC ((M /. (T2 . t)),(T2 . t))),S)) . m ) ) ; :: thesis: T1 = T2
defpred S1[ Element of NAT ] means (dom T1) -level $1 = (dom T2) -level $1;
A17: for n being Element of NAT st S1[n] holds
S1[n + 1]
proof
let n be Element of NAT ; :: thesis: ( S1[n] implies S1[n + 1] )
assume A18: S1[n] ; :: thesis: S1[n + 1]
set U2 = { (succ w) where w is Element of dom T2 : len w = n } ;
set U1 = { (succ w) where w is Element of dom T1 : len w = n } ;
A19: (dom T2) -level n = { v where v is Element of dom T2 : len v = n } by TREES_2:def 6;
A20: (dom T1) -level n = { v where v is Element of dom T1 : len v = n } by TREES_2:def 6;
A21: union { (succ w) where w is Element of dom T1 : len w = n } = union { (succ w) where w is Element of dom T2 : len w = n }
proof
hereby :: according to TARSKI:def 3,XBOOLE_0:def 10 :: thesis: union { (succ w) where w is Element of dom T2 : len w = n } c= union { (succ w) where w is Element of dom T1 : len w = n }
let a be set ; :: thesis: ( a in union { (succ w) where w is Element of dom T1 : len w = n } implies a in union { (succ w) where w is Element of dom T2 : len w = n } )
assume a in union { (succ w) where w is Element of dom T1 : len w = n } ; :: thesis: a in union { (succ w) where w is Element of dom T2 : len w = n }
then consider A being set such that
A22: a in A and
A23: A in { (succ w) where w is Element of dom T1 : len w = n } by TARSKI:def 4;
consider w being Element of dom T1 such that
A24: A = succ w and
A25: len w = n by A23;
w in (dom T1) -level n by A20, A25;
then consider v being Element of dom T2 such that
A26: w = v and
A27: len v = n by A18, A19;
A28: w = w | (Seg (len w)) by FINSEQ_3:49;
defpred S2[ Element of NAT ] means ( $1 <= len w & w | (Seg $1) in dom T1 & v | (Seg $1) in dom T2 implies T1 . (w | (Seg $1)) = T2 . (w | (Seg $1)) );
A29: for n being Element of NAT st S2[n] holds
S2[n + 1]
proof
let n be Element of NAT ; :: thesis: ( S2[n] implies S2[n + 1] )
assume that
A30: S2[n] and
A31: n + 1 <= len w and
A32: w | (Seg (n + 1)) in dom T1 and
A33: v | (Seg (n + 1)) in dom T2 ; :: thesis: T1 . (w | (Seg (n + 1))) = T2 . (w | (Seg (n + 1)))
set t1 = w | (Seg n);
A34: 1 <= n + 1 by NAT_1:11;
A35: len (w | (Seg (n + 1))) = n + 1 by A31, FINSEQ_1:17;
then len (w | (Seg (n + 1))) in Seg (n + 1) by A34, FINSEQ_1:1;
then A36: w . (n + 1) = (w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))) by A35, FUNCT_1:49;
n + 1 in dom w by A31, A34, FINSEQ_3:25;
then A37: w | (Seg (n + 1)) = (w | (Seg n)) ^ <*((w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))))*> by A36, FINSEQ_5:10;
A38: n <= n + 1 by NAT_1:11;
then A39: Seg n c= Seg (n + 1) by FINSEQ_1:5;
then v | (Seg n) = (v | (Seg (n + 1))) | (Seg n) by RELAT_1:74;
then A40: v | (Seg n) is_a_prefix_of v | (Seg (n + 1)) by TREES_1:def 1;
w | (Seg n) = (w | (Seg (n + 1))) | (Seg n) by A39, RELAT_1:74;
then w | (Seg n) is_a_prefix_of w | (Seg (n + 1)) by TREES_1:def 1;
then reconsider t1 = w | (Seg n) as Element of dom T1 by A32, TREES_1:20;
reconsider t2 = t1 as Element of dom T2 by A26, A33, A40, TREES_1:20;
A41: succ t1 = { (t1 ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T1 . t1)),(T1 . t1))) } by A14;
A42: (w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))) is Element of NAT by ORDINAL1:def 12;
then t1 ^ <*((w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))))*> in succ t1 by A32, A37, TREES_2:12;
then consider k being Element of NAT such that
A43: t1 ^ <*((w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))))*> = t1 ^ <*k*> and
A44: k in card (NIC ((M /. (T1 . t1)),(T1 . t1))) by A41;
A45: (w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))) in card (NIC ((M /. (T2 . t2)),(T2 . t2))) by A30, A31, A33, A38, A40, A43, A44, FINSEQ_2:17, TREES_1:20, XXREAL_0:2;
k = (w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))) by A43, FINSEQ_2:17;
hence T1 . (w | (Seg (n + 1))) = (LocSeq ((NIC ((M /. (T1 . t1)),(T1 . t1))),S)) . ((w | (Seg (n + 1))) . (len (w | (Seg (n + 1))))) by A14, A37, A44
.= T2 . (w | (Seg (n + 1))) by A16, A30, A31, A33, A38, A40, A42, A37, A45, TREES_1:20, XXREAL_0:2 ;
:: thesis: verum
end;
A46: S2[ 0 ] by A13, A15;
for n being Element of NAT holds S2[n] from NAT_1:sch 1(A46, A29);
then A47: T1 . w = T2 . w by A26, A28;
A48: succ v in { (succ w) where w is Element of dom T2 : len w = n } by A27;
( succ v = { (v ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T2 . v)),(T2 . v))) } & succ w = { (w ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T1 . w)),(T1 . w))) } ) by A14, A16;
hence a in union { (succ w) where w is Element of dom T2 : len w = n } by A22, A24, A26, A47, A48, TARSKI:def 4; :: thesis: verum
end;
let a be set ; :: according to TARSKI:def 3 :: thesis: ( not a in union { (succ w) where w is Element of dom T2 : len w = n } or a in union { (succ w) where w is Element of dom T1 : len w = n } )
assume a in union { (succ w) where w is Element of dom T2 : len w = n } ; :: thesis: a in union { (succ w) where w is Element of dom T1 : len w = n }
then consider A being set such that
A49: a in A and
A50: A in { (succ w) where w is Element of dom T2 : len w = n } by TARSKI:def 4;
consider w being Element of dom T2 such that
A51: A = succ w and
A52: len w = n by A50;
w in (dom T2) -level n by A19, A52;
then consider v being Element of dom T1 such that
A53: w = v and
A54: len v = n by A18, A20;
A55: w = w | (Seg (len w)) by FINSEQ_3:49;
defpred S2[ Element of NAT ] means ( $1 <= len w & w | (Seg $1) in dom T1 & v | (Seg $1) in dom T2 implies T1 . (w | (Seg $1)) = T2 . (w | (Seg $1)) );
A56: for n being Element of NAT st S2[n] holds
S2[n + 1]
proof
let n be Element of NAT ; :: thesis: ( S2[n] implies S2[n + 1] )
assume that
A57: S2[n] and
A58: n + 1 <= len w and
A59: w | (Seg (n + 1)) in dom T1 and
A60: v | (Seg (n + 1)) in dom T2 ; :: thesis: T1 . (w | (Seg (n + 1))) = T2 . (w | (Seg (n + 1)))
set t1 = w | (Seg n);
A61: 1 <= n + 1 by NAT_1:11;
A62: len (w | (Seg (n + 1))) = n + 1 by A58, FINSEQ_1:17;
then len (w | (Seg (n + 1))) in Seg (n + 1) by A61, FINSEQ_1:1;
then A63: w . (n + 1) = (w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))) by A62, FUNCT_1:49;
n + 1 in dom w by A58, A61, FINSEQ_3:25;
then A64: w | (Seg (n + 1)) = (w | (Seg n)) ^ <*((w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))))*> by A63, FINSEQ_5:10;
A65: n <= n + 1 by NAT_1:11;
then A66: Seg n c= Seg (n + 1) by FINSEQ_1:5;
then v | (Seg n) = (v | (Seg (n + 1))) | (Seg n) by RELAT_1:74;
then A67: v | (Seg n) is_a_prefix_of v | (Seg (n + 1)) by TREES_1:def 1;
w | (Seg n) = (w | (Seg (n + 1))) | (Seg n) by A66, RELAT_1:74;
then w | (Seg n) is_a_prefix_of w | (Seg (n + 1)) by TREES_1:def 1;
then reconsider t1 = w | (Seg n) as Element of dom T1 by A59, TREES_1:20;
reconsider t2 = t1 as Element of dom T2 by A53, A60, A67, TREES_1:20;
A68: succ t1 = { (t1 ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T1 . t1)),(T1 . t1))) } by A14;
A69: (w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))) is Element of NAT by ORDINAL1:def 12;
then t1 ^ <*((w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))))*> in succ t1 by A59, A64, TREES_2:12;
then consider k being Element of NAT such that
A70: t1 ^ <*((w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))))*> = t1 ^ <*k*> and
A71: k in card (NIC ((M /. (T1 . t1)),(T1 . t1))) by A68;
A72: (w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))) in card (NIC ((M /. (T2 . t2)),(T2 . t2))) by A57, A58, A60, A65, A67, A70, A71, FINSEQ_2:17, TREES_1:20, XXREAL_0:2;
k = (w | (Seg (n + 1))) . (len (w | (Seg (n + 1)))) by A70, FINSEQ_2:17;
hence T1 . (w | (Seg (n + 1))) = (LocSeq ((NIC ((M /. (T1 . t1)),(T1 . t1))),S)) . ((w | (Seg (n + 1))) . (len (w | (Seg (n + 1))))) by A14, A64, A71
.= T2 . (w | (Seg (n + 1))) by A16, A57, A58, A60, A65, A67, A69, A64, A72, TREES_1:20, XXREAL_0:2 ;
:: thesis: verum
end;
A73: S2[ 0 ] by A13, A15;
for n being Element of NAT holds S2[n] from NAT_1:sch 1(A73, A56);
then A74: T1 . w = T2 . w by A53, A55;
A75: succ v in { (succ w) where w is Element of dom T1 : len w = n } by A54;
( succ v = { (v ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T1 . v)),(T1 . v))) } & succ w = { (w ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T2 . w)),(T2 . w))) } ) by A14, A16;
hence a in union { (succ w) where w is Element of dom T1 : len w = n } by A49, A51, A53, A74, A75, TARSKI:def 4; :: thesis: verum
end;
(dom T1) -level (n + 1) = union { (succ w) where w is Element of dom T1 : len w = n } by TREES_9:45;
hence S1[n + 1] by A21, TREES_9:45; :: thesis: verum
end;
(dom T1) -level 0 = {{}} by TREES_9:44
.= (dom T2) -level 0 by TREES_9:44 ;
then A76: S1[ 0 ] ;
A77: for n being Element of NAT holds S1[n] from NAT_1:sch 1(A76, A17);
for p being FinSequence of NAT st p in dom T1 holds
T1 . p = T2 . p
proof
let p be FinSequence of NAT ; :: thesis: ( p in dom T1 implies T1 . p = T2 . p )
defpred S2[ Element of NAT ] means ( $1 <= len p & p | (Seg $1) in dom T1 implies T1 . (p | (Seg $1)) = T2 . (p | (Seg $1)) );
A78: p | (Seg (len p)) = p by FINSEQ_3:49;
A79: for n being Element of NAT st S2[n] holds
S2[n + 1]
proof
let n be Element of NAT ; :: thesis: ( S2[n] implies S2[n + 1] )
assume that
A80: S2[n] and
A81: n + 1 <= len p and
A82: p | (Seg (n + 1)) in dom T1 ; :: thesis: T1 . (p | (Seg (n + 1))) = T2 . (p | (Seg (n + 1)))
set t1 = p | (Seg n);
A83: 1 <= n + 1 by NAT_1:11;
A84: len (p | (Seg (n + 1))) = n + 1 by A81, FINSEQ_1:17;
then len (p | (Seg (n + 1))) in Seg (n + 1) by A83, FINSEQ_1:1;
then A85: p . (n + 1) = (p | (Seg (n + 1))) . (len (p | (Seg (n + 1)))) by A84, FUNCT_1:49;
n + 1 in dom p by A81, A83, FINSEQ_3:25;
then A86: p | (Seg (n + 1)) = (p | (Seg n)) ^ <*((p | (Seg (n + 1))) . (len (p | (Seg (n + 1)))))*> by A85, FINSEQ_5:10;
A87: n <= n + 1 by NAT_1:11;
then Seg n c= Seg (n + 1) by FINSEQ_1:5;
then p | (Seg n) = (p | (Seg (n + 1))) | (Seg n) by RELAT_1:74;
then p | (Seg n) is_a_prefix_of p | (Seg (n + 1)) by TREES_1:def 1;
then reconsider t1 = p | (Seg n) as Element of dom T1 by A82, TREES_1:20;
reconsider t2 = t1 as Element of dom T2 by A77, TREES_2:38;
A88: succ t1 = { (t1 ^ <*k*>) where k is Element of NAT : k in card (NIC ((M /. (T1 . t1)),(T1 . t1))) } by A14;
A89: (p | (Seg (n + 1))) . (len (p | (Seg (n + 1)))) is Element of NAT by ORDINAL1:def 12;
then t1 ^ <*((p | (Seg (n + 1))) . (len (p | (Seg (n + 1)))))*> in succ t1 by A82, A86, TREES_2:12;
then consider k being Element of NAT such that
A90: t1 ^ <*((p | (Seg (n + 1))) . (len (p | (Seg (n + 1)))))*> = t1 ^ <*k*> and
A91: k in card (NIC ((M /. (T1 . t1)),(T1 . t1))) by A88;
A92: (p | (Seg (n + 1))) . (len (p | (Seg (n + 1)))) in card (NIC ((M /. (T2 . t2)),(T2 . t2))) by A80, A81, A87, A90, A91, FINSEQ_2:17, XXREAL_0:2;
k = (p | (Seg (n + 1))) . (len (p | (Seg (n + 1)))) by A90, FINSEQ_2:17;
hence T1 . (p | (Seg (n + 1))) = (LocSeq ((NIC ((M /. (T1 . t1)),(T1 . t1))),S)) . ((p | (Seg (n + 1))) . (len (p | (Seg (n + 1))))) by A14, A86, A91
.= T2 . (p | (Seg (n + 1))) by A16, A80, A81, A87, A89, A86, A92, XXREAL_0:2 ;
:: thesis: verum
end;
A93: S2[ 0 ] by A13, A15;
for n being Element of NAT holds S2[n] from NAT_1:sch 1(A93, A79);
hence ( p in dom T1 implies T1 . p = T2 . p ) by A78; :: thesis: verum
end;
hence T1 = T2 by A77, TREES_2:31, TREES_2:38; :: thesis: verum