defpred S1[ Nat] means for p, q being Element of $1 -tuples_on NAT st p <> q holds
ex i being Element of NAT st
( i in Seg $1 & p . i <> q . i & ( for k being Element of NAT st 1 <= k & k < i holds
p . k = q . k ) );
A1:
S1[ 0 ]
A3:
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 A4:
for
p,
q being
Element of
n -tuples_on NAT st
p <> q holds
ex
i being
Element of
NAT st
(
i in Seg n &
p . i <> q . i & ( for
k being
Element of
NAT st 1
<= k &
k < i holds
p . k = q . k ) )
;
:: thesis: S1[n + 1]
let p,
q be
Element of
(n + 1) -tuples_on NAT ;
:: thesis: ( p <> q implies ex i being Element of NAT st
( i in Seg (n + 1) & p . i <> q . i & ( for k being Element of NAT st 1 <= k & k < i holds
p . k = q . k ) ) )
consider t1 being
Element of
n -tuples_on NAT ,
d1 being
Element of
NAT such that A5:
p = t1 ^ <*d1*>
by FINSEQ_2:137;
consider t2 being
Element of
n -tuples_on NAT ,
d2 being
Element of
NAT such that A6:
q = t2 ^ <*d2*>
by FINSEQ_2:137;
assume A7:
p <> q
;
:: thesis: ex i being Element of NAT st
( i in Seg (n + 1) & p . i <> q . i & ( for k being Element of NAT st 1 <= k & k < i holds
p . k = q . k ) )
A8:
(
len t1 = n &
len t2 = n )
by FINSEQ_1:def 18;
per cases
( t1 <> t2 or t1 = t2 )
;
suppose
t1 <> t2
;
:: thesis: ex i being Element of NAT st
( i in Seg (n + 1) & p . i <> q . i & ( for k being Element of NAT st 1 <= k & k < i holds
p . k = q . k ) )then consider i being
Element of
NAT such that A9:
i in Seg n
and A10:
t1 . i <> t2 . i
and A11:
for
k being
Element of
NAT st 1
<= k &
k < i holds
t1 . k = t2 . k
by A4;
take
i
;
:: thesis: ( i in Seg (n + 1) & p . i <> q . i & ( for k being Element of NAT st 1 <= k & k < i holds
p . k = q . k ) )thus
i in Seg (n + 1)
by A9, FINSEQ_2:9;
:: thesis: ( p . i <> q . i & ( for k being Element of NAT st 1 <= k & k < i holds
p . k = q . k ) )
(
i in dom t1 &
i in dom t2 )
by A8, A9, FINSEQ_1:def 3;
then
(
p . i = t1 . i &
q . i = t2 . i )
by A5, A6, FINSEQ_1:def 7;
hence
p . i <> q . i
by A10;
:: thesis: for k being Element of NAT st 1 <= k & k < i holds
p . k = q . klet k be
Element of
NAT ;
:: thesis: ( 1 <= k & k < i implies p . k = q . k )assume that A12:
1
<= k
and A13:
k < i
;
:: thesis: p . k = q . kA14:
t1 . k = t2 . k
by A11, A12, A13;
i <= n
by A9, FINSEQ_1:3;
then
k <= n
by A13, XXREAL_0:2;
then A15:
(
k in dom t1 &
k in dom t2 )
by A8, A12, FINSEQ_3:27;
hence p . k =
t2 . k
by A5, A14, FINSEQ_1:def 7
.=
q . k
by A6, A15, FINSEQ_1:def 7
;
:: thesis: verum end; suppose A16:
t1 = t2
;
:: thesis: ex i being Element of NAT st
( i in Seg (n + 1) & p . i <> q . i & ( for k being Element of NAT st 1 <= k & k < i holds
p . k = q . k ) )take i =
n + 1;
:: thesis: ( i in Seg (n + 1) & p . i <> q . i & ( for k being Element of NAT st 1 <= k & k < i holds
p . k = q . k ) )thus
i in Seg (n + 1)
by FINSEQ_1:6;
:: thesis: ( p . i <> q . i & ( for k being Element of NAT st 1 <= k & k < i holds
p . k = q . k ) )
(
p . i = d1 &
q . i = d2 )
by A5, A6, FINSEQ_2:136;
hence
p . i <> q . i
by A5, A6, A7, A16;
:: thesis: for k being Element of NAT st 1 <= k & k < i holds
p . k = q . klet k be
Element of
NAT ;
:: thesis: ( 1 <= k & k < i implies p . k = q . k )assume that A17:
1
<= k
and A18:
k < i
;
:: thesis: p . k = q . k
k <= n
by A18, NAT_1:13;
then A19:
(
k in dom t1 &
k in dom t2 )
by A8, A17, FINSEQ_3:27;
hence p . k =
t2 . k
by A5, A16, FINSEQ_1:def 7
.=
q . k
by A6, A19, FINSEQ_1:def 7
;
:: thesis: verum end; end;
end;
thus
for n being Element of NAT holds S1[n]
from NAT_1:sch 1(A1, A3); :: thesis: verum