:: CALCUL_2 semantic presentation
:: deftheorem defines seq CALCUL_2:def 1 :
theorem E1: :: CALCUL_2:1
theorem E2: :: CALCUL_2:2
theorem E3: :: CALCUL_2:3
theorem E4: :: CALCUL_2:4
theorem E5: :: CALCUL_2:5
theorem E6: :: CALCUL_2:6
proof
let c
1, c
2 be
Nat;
defpred S
1[
Nat] means
seq c
1,a
1,a
1 are_equipotent ;
E7:
S
1[0]
by E2;
E8:
for b
1 being
Nat holds
( S
1[b
1] implies S
1[b
1 + 1] )
proof
let c
3 be
Nat;
assume E9:
seq c
1,c
3,c
3 are_equipotent
;
reconsider c
4 = c
1 + c
3 as
Nat ;
E10:
seq c
1,
(c3 + 1) = (seq c1,c3) \/ {(c4 + 1)}
by E5;
E11: c
3 + 1 =
succ c
3
by CARD_1:52
.=
c
3 \/ {c3}
by ORDINAL1:def 1
;
E12:
{(c4 + 1)},
{c3} are_equipotent
by CARD_1:48;
hence
S
1[c
3 + 1]
by E9, E10, E11, E12, E13, CARD_1:58;
end;
for b
1 being
Nat holds
S
1[b
1]
from NAT_1:sch 1(E7, E8);
hence
seq c
1,c
2,c
2 are_equipotent
;
end;
theorem E7: :: CALCUL_2:7
theorem :: CALCUL_2:8
theorem :: CALCUL_2:9
theorem E8: :: CALCUL_2:10
theorem E9: :: CALCUL_2:11
theorem E10: :: CALCUL_2:12
theorem E11: :: CALCUL_2:13
proof
let c
1 be
Nat;
let c
2, c
3 be
FinSequence of
CQC-WFF ;
assume
c
1 in dom (Sgm (seq (len c2),(len c3)))
;
then
c
1 in dom c
3
by E9;
then E12:
( 1
<= c
1 & c
1 <= len c
3 )
by FINSEQ_3:27;
defpred S
1[
Nat] means ( ( 1
<= a
1 & a
1 <= len c
3 ) implies ( ( for b
1 being
Nat holds
( ( 1
<= b
1 & b
1 <= a
1 ) implies (
(Sgm (seq (len c2),(len c3))) . b
1 = (len c2) + b
1 ) ) ) ) );
E13:
S
1[0]
;
E14:
for b
1 being
Nat holds
( S
1[b
1] implies S
1[b
1 + 1] )
proof
let c
4 be
Nat;
assume E15:
S
1[c
4]
;
assume E16:
( 1
<= c
4 + 1 & c
4 + 1
<= len c
3 )
;
let c
5 be
Nat;
assume E17:
( 1
<= c
5 & c
5 <= c
4 + 1 )
;
E18:
( c
5 <= c
4 implies
(Sgm (seq (len c2),(len c3))) . c
5 = (len c2) + c
5 )
by E15, E16, E17, XXREAL_0:2, NAT_1:38;
now
assume E19:
c
5 = c
4 + 1
;
dom (Sgm (seq (len c2),(len c3))) = dom c
3
by E9;
then
c
5 in dom (Sgm (seq (len c2),(len c3)))
by E16, E19, FINSEQ_3:27;
then
(Sgm (seq (len c2),(len c3))) . c
5 in rng (Sgm (seq (len c2),(len c3)))
by FUNCT_1:12;
then reconsider c
6 =
(Sgm (seq (len c2),(len c3))) . c
5 as
Nat ;
E20:
now
assume E21:
c
6 < (len c2) + (c4 + 1)
;
now
assume E23:
c
4 <> 0
;
E24:
c
6 < ((len c2) + c4) + 1
by E21;
0
< c
4
by E23;
then E25:
( 0
+ 1
<= c
4 & c
4 <= c
4 + 1 )
by NAT_1:38;
E26:
(Sgm (seq (len c2),(len c3))) . c
4 = (len c2) + c
4
by E15, E16, E25, XXREAL_0:2;
then reconsider c
7 =
(Sgm (seq (len c2),(len c3))) . c
4 as
Nat ;
E27:
( 1
<= c
4 & c
4 < c
4 + 1 & c
4 + 1
<= len (Sgm (seq (len c2),(len c3))) )
by E16, E25, E8, NAT_1:38;
( c
6 <= c
7 &
seq (len c2),
(len c3) c= Seg ((len c2) + (len c3)) )
by E24, E26, E7, NAT_1:38;
hence
not verum
by E19, E27, FINSEQ_1:def 13;
end;
hence
not verum
by E22;
end;
now
assume E21:
c
6 > (len c2) + (c4 + 1)
;
E22:
1
+ (len c2) <= (1 + (len c2)) + c
4
by NAT_1:29;
(len c2) + (c4 + 1) <= (len c3) + (len c2)
by E16, XREAL_1:8;
then
(len c2) + (c4 + 1) in seq (len c2),
(len c3)
by E22;
then
(len c2) + (c4 + 1) in rng (Sgm (seq (len c2),(len c3)))
by E10;
then consider c
7 being
Nat such that E23:
( c
7 in dom (Sgm (seq (len c2),(len c3))) &
(Sgm (seq (len c2),(len c3))) . c
7 = (len c2) + (c4 + 1) )
by FINSEQ_2:11;
E24:
now
assume E25:
c
7 <= c
4 + 1
;
hence
not verum
by E19, E21, E23, E25, NAT_1:26;
end;
reconsider c
8 =
(Sgm (seq (len c2),(len c3))) . c
7 as
Nat by E23;
E25:
( 1
<= c
5 & c
5 < c
7 & c
7 <= len (Sgm (seq (len c2),(len c3))) )
by E19, E23, E24, FINSEQ_3:27, NAT_1:29;
seq (len c2),
(len c3) c= Seg ((len c2) + (len c3))
by E7;
hence
not verum
by E21, E23, E25, FINSEQ_1:def 13;
end;
hence
(Sgm (seq (len c2),(len c3))) . c
5 = (len c2) + c
5
by E19, E20, XXREAL_0:1;
end;
hence
(Sgm (seq (len c2),(len c3))) . c
5 = (len c2) + c
5
by E17, E18, NAT_1:26;
end;
for b
1 being
Nat holds
S
1[b
1]
from NAT_1:sch 1(E13, E14);
hence
(Sgm (seq (len c2),(len c3))) . c
1 = (len c2) + c
1
by E12;
end;
theorem E12: :: CALCUL_2:14
theorem E13: :: CALCUL_2:15
theorem E14: :: CALCUL_2:16
theorem E15: :: CALCUL_2:17
proof
let c
1, c
2 be
FinSequence of
CQC-WFF ;
rng (Sgm (seq (len c1),(len c2))) = seq (len c1),
(len c2)
by E10;
then E16:
rng (Sgm (seq (len c1),(len c2))) c= dom (c1 ^ c2)
by E12;
dom (Seq ((c1 ^ c2) | (seq (len c1),(len c2)))) = dom ((Sgm (seq (len c1),(len c2))) * (c1 ^ c2))
by E14;
then
dom (Seq ((c1 ^ c2) | (seq (len c1),(len c2)))) = dom (Sgm (seq (len c1),(len c2)))
by E16, RELAT_1:46;
hence
dom (Seq ((c1 ^ c2) | (seq (len c1),(len c2)))) = dom c
2
by E9;
end;
theorem E16: :: CALCUL_2:18
proof
let c
1, c
2 be
FinSequence of
CQC-WFF ;
E17:
for b
1 being
Nat holds
( b
1 in dom (Seq ((c2 ^ c1) | (seq (len c2),(len c1)))) implies
(Seq ((c2 ^ c1) | (seq (len c2),(len c1)))) . b
1 = c
1 . b
1 )
proof
let c
3 be
Nat;
assume E18:
c
3 in dom (Seq ((c2 ^ c1) | (seq (len c2),(len c1))))
;
E19:
(Seq ((c2 ^ c1) | (seq (len c2),(len c1)))) . c
3 = ((Sgm (seq (len c2),(len c1))) * (c2 ^ c1)) . c
3
by E14;
E20:
c
3 in dom c
1
by E18, E15;
then E21:
c
3 in dom (Sgm (seq (len c2),(len c1)))
by E9;
then
(Seq ((c2 ^ c1) | (seq (len c2),(len c1)))) . c
3 = (c2 ^ c1) . ((Sgm (seq (len c2),(len c1))) . c3)
by E19, FUNCT_1:23;
then
(Seq ((c2 ^ c1) | (seq (len c2),(len c1)))) . c
3 = (c2 ^ c1) . ((len c2) + c3)
by E21, E11;
hence
(Seq ((c2 ^ c1) | (seq (len c2),(len c1)))) . c
3 = c
1 . c
3
by E20, FINSEQ_1:def 7;
end;
dom (Seq ((c2 ^ c1) | (seq (len c2),(len c1)))) = dom c
1
by E15;
then
Seq ((c2 ^ c1) | (seq (len c2),(len c1))) = c
1
by E17, FINSEQ_1:17;
hence
c
1 is_Subsequence_of c
2 ^ c
1
by CALCUL_1:def 4;
end;
:: deftheorem defines Per CALCUL_2:def 2 :
theorem E17: :: CALCUL_2:19
theorem E18: :: CALCUL_2:20
:: deftheorem E19 defines Begin CALCUL_2:def 3 :
definition