:: FINSEQ_3 semantic presentation
theorem Th1: :: FINSEQ_3:1
theorem Th2: :: FINSEQ_3:2
theorem Th3: :: FINSEQ_3:3
theorem Th4: :: FINSEQ_3:4
theorem Th5: :: FINSEQ_3:5
theorem Th6: :: FINSEQ_3:6
theorem Th7: :: FINSEQ_3:7
theorem Th8: :: FINSEQ_3:8
canceled;
theorem Th9: :: FINSEQ_3:9
theorem Th10: :: FINSEQ_3:10
theorem Th11: :: FINSEQ_3:11
theorem Th12: :: FINSEQ_3:12
theorem Th13: :: FINSEQ_3:13
theorem Th14: :: FINSEQ_3:14
theorem Th15: :: FINSEQ_3:15
theorem Th16: :: FINSEQ_3:16
theorem Th17: :: FINSEQ_3:17
theorem Th18: :: FINSEQ_3:18
theorem Th19: :: FINSEQ_3:19
theorem Th20: :: FINSEQ_3:20
theorem Th21: :: FINSEQ_3:21
canceled;
theorem Th22: :: FINSEQ_3:22
theorem Th23: :: FINSEQ_3:23
theorem Th24: :: FINSEQ_3:24
theorem Th25: :: FINSEQ_3:25
theorem Th26: :: FINSEQ_3:26
theorem Th27: :: FINSEQ_3:27
theorem Th28: :: FINSEQ_3:28
theorem Th29: :: FINSEQ_3:29
theorem Th30: :: FINSEQ_3:30
theorem Th31: :: FINSEQ_3:31
theorem Th32: :: FINSEQ_3:32
theorem Th33: :: FINSEQ_3:33
theorem Th34: :: FINSEQ_3:34
theorem Th35: :: FINSEQ_3:35
canceled;
theorem Th36: :: FINSEQ_3:36
canceled;
theorem Th37: :: FINSEQ_3:37
canceled;
theorem Th38: :: FINSEQ_3:38
theorem Th39: :: FINSEQ_3:39
theorem Th40: :: FINSEQ_3:40
theorem Th41: :: FINSEQ_3:41
theorem Th42: :: FINSEQ_3:42
theorem Th43: :: FINSEQ_3:43
Lemma63:
for A being set
for k being natural number st A c= Seg k holds
Sgm A is one-to-one
theorem Th44: :: FINSEQ_3:44
theorem Th45: :: FINSEQ_3:45
theorem Th46: :: FINSEQ_3:46
theorem Th47: :: FINSEQ_3:47
canceled;
theorem Th48: :: FINSEQ_3:48
theorem Th49: :: FINSEQ_3:49
theorem Th50: :: FINSEQ_3:50
theorem Th51: :: FINSEQ_3:51
theorem Th52: :: FINSEQ_3:52
Lemma149:
for k being natural number holds (Sgm (Seg (k + 0))) | (Seg k) = Sgm (Seg k)
E150:
now
let n be
natural number ;
assume E40:
for
k being
natural number holds
(Sgm (Seg (k + n))) | (Seg k) = Sgm (Seg k)
;
let k be
natural number ;
set X =
Sgm (Seg (k + (n + 1)));
set Y =
Sgm (Seg (k + 1));
Sgm (Seg (k + (n + 1))) = Sgm (Seg ((k + 1) + n))
;
then E42:
(Sgm (Seg (k + (n + 1)))) | (Seg (k + 1)) = Sgm (Seg (k + 1))
by ;
(Sgm (Seg (k + 1))) | (Seg k) = Sgm (Seg k)
proof
reconsider p =
(Sgm (Seg (k + 1))) | (Seg k) as
FinSequence of
NAT by FINSEQ_1:23;
E44:
rng (Sgm (Seg (k + 1))) = Seg (k + 1)
by FINSEQ_1:def 13;
E45:
Sgm (Seg (k + 1)) is
one-to-one
by ;
E50:
len (Sgm (Seg (k + 1))) = k + 1
by ;
then E51:
(
dom (Sgm (Seg (k + 1))) = Seg (k + 1) &
k <= k + 1 )
by FINSEQ_1:def 3, NAT_1:37;
then E52:
dom p = Seg k
by Th2, FINSEQ_1:21;
E70:
(Sgm (Seg (k + 1))) . (k + 1) = k + 1
proof
assume E71:
not
(Sgm (Seg (k + 1))) . (k + 1) = k + 1
;
k + 1
in dom (Sgm (Seg (k + 1)))
by Th3, FINSEQ_1:6;
then
(
(Sgm (Seg (k + 1))) . (k + 1) in Seg (k + 1) & not
(Sgm (Seg (k + 1))) . (k + 1) in {(k + 1)} )
by , Th9, FUNCT_1:def 5, TARSKI:def 1;
then
(Sgm (Seg (k + 1))) . (k + 1) in (Seg (k + 1)) \ {(k + 1)}
by XBOOLE_0:def 4;
then E72:
(Sgm (Seg (k + 1))) . (k + 1) in Seg k
by FINSEQ_1:12;
then reconsider n =
(Sgm (Seg (k + 1))) . (k + 1) as
Element of
NAT ;
k + 1
in rng (Sgm (Seg (k + 1)))
by , FINSEQ_1:6;
then consider x being
set such that E73:
x in dom (Sgm (Seg (k + 1)))
and E74:
(Sgm (Seg (k + 1))) . x = k + 1
by FUNCT_1:def 5;
reconsider x =
x as
Element of
NAT by Th11;
E75:
( 1
<= x &
x <= k + 1 )
by Th2, Th11, ;
then
x < k + 1
by Th13, REAL_1:def 5;
then
(
k + 1
< n &
n <= k &
k < k + 1 )
by Th2, E40, E42, Th13, FINSEQ_1:3, FINSEQ_1:def 13, XREAL_1:31;
hence
contradiction
by XXREAL_0:2;
end;
rng p = (rng (Sgm (Seg (k + 1)))) \ {((Sgm (Seg (k + 1))) . (k + 1))}
proof
thus
rng p c= (rng (Sgm (Seg (k + 1)))) \ {((Sgm (Seg (k + 1))) . (k + 1))}
:: according to XBOOLE_0:def 10
proof
let x be
set ;
:: according to TARSKI:def 3
assume E76:
x in rng p
;
E77:
rng p c= rng (Sgm (Seg (k + 1)))
by RELAT_1:99;
now
assume E78:
x in {(k + 1)}
;
consider y being
set such that E79:
y in dom p
and E80:
p . y = x
by Th9, FUNCT_1:def 5;
reconsider y =
y as
Element of
NAT by E42;
Seg k c= Seg (k + 1)
by ;
then E83:
(
y in dom (Sgm (Seg (k + 1))) &
(Sgm (Seg (k + 1))) . y = p . y &
x = k + 1 &
k + 1
in dom (Sgm (Seg (k + 1))) )
by Th3, Th4, Th11, E42, FINSEQ_1:6, FUNCT_1:70, TARSKI:def 1;
(
y <= k &
k < k + 1 )
by Th4, E42, FINSEQ_1:3, XREAL_1:31;
hence
contradiction
by Th1, Th5, Th13, E44, FUNCT_1:def 8;
end;
hence
x in (rng (Sgm (Seg (k + 1)))) \ {((Sgm (Seg (k + 1))) . (k + 1))}
by Th5, Th9, E40, XBOOLE_0:def 4;
end;
let x be
set ;
:: according to TARSKI:def 3
assume E84:
x in (rng (Sgm (Seg (k + 1)))) \ {((Sgm (Seg (k + 1))) . (k + 1))}
;
then
x in rng (Sgm (Seg (k + 1)))
;
then consider y being
set such that E85:
y in dom (Sgm (Seg (k + 1)))
and E86:
(Sgm (Seg (k + 1))) . y = x
by FUNCT_1:def 5;
then
y in (Seg (k + 1)) \ {(k + 1)}
by Th3, E40, XBOOLE_0:def 4;
then E87:
y in dom p
by Th4, FINSEQ_1:12;
then
p . y = (Sgm (Seg (k + 1))) . y
by FUNCT_1:70;
hence
x in rng p
by Th11, E42, FUNCT_1:def 5;
end;
then E88:
rng p = Seg k
by , Th5, FINSEQ_1:12;
now
let i be
natural number ;
let j be
natural number ;
let l be
natural number ;
let m be
natural number ;
assume that E89:
( 1
<= i &
i < j &
j <= len p )
and E90:
(
l = p . i &
m = p . j )
;
E91:
(
len p = k &
i <= len p & 1
<= j )
by Th2, Th3, E40, XXREAL_0:2, FINSEQ_1:21;
then
(
i in dom p &
j in dom p &
len (Sgm (Seg (k + 1))) = k + 1 &
k <= k + 1 )
by Th4, E40, , FINSEQ_1:3, NAT_1:37;
then
(
p . i = (Sgm (Seg (k + 1))) . i &
p . j = (Sgm (Seg (k + 1))) . j &
j <= len (Sgm (Seg (k + 1))) )
by E40, E42, XXREAL_0:2, FUNCT_1:70;
hence
l < m
by E40, Th11, FINSEQ_1:def 13;
end;
hence
(Sgm (Seg (k + 1))) | (Seg k) = Sgm (Seg k)
by Th9, FINSEQ_1:def 13;
end;
then
(
Sgm (Seg k) = (Sgm (Seg (k + (n + 1)))) | ((Seg (k + 1)) /\ (Seg k)) &
k <= k + 1 )
by , NAT_1:37, RELAT_1:100;
hence
(Sgm (Seg (k + (n + 1)))) | (Seg k) = Sgm (Seg k)
by FINSEQ_1:9;
end;
Lemma151:
for n, k being natural number holds (Sgm (Seg (k + n))) | (Seg k) = Sgm (Seg k)
theorem Th53: :: FINSEQ_3:53
E152:
now
let k be
Element of
NAT ;
assume E40:
Sgm (Seg k) = idseq k
;
E42:
len (idseq (k + 1)) = k + 1
by FINSEQ_2:55;
then E44:
len (Sgm (Seg (k + 1))) = len (idseq (k + 1))
by ;
now
let j be
Element of
NAT ;
assume E45:
j in Seg (k + 1)
;
then E50:
j in (Seg k) \/ {(k + 1)}
by FINSEQ_1:11;
now
per cases
( j in Seg k or j in {(k + 1)} )
by Th4, XBOOLE_0:def 2;
suppose
j in {(k + 1)}
;
then E52:
j = k + 1
by TARSKI:def 1;
then E70:
j in Seg (k + 1)
by FINSEQ_1:6;
set X =
Sgm (Seg (k + 1));
set Y =
Sgm (Seg k);
now
assume
(Sgm (Seg (k + 1))) . j <> j
;
then E71:
( not
(Sgm (Seg (k + 1))) . j in {j} &
Seg (k + 1) c= Seg (k + 1) )
by TARSKI:def 1;
E72:
(
rng (Sgm (Seg (k + 1))) = Seg (k + 1) &
dom (Sgm (Seg (k + 1))) = Seg (k + 1) )
by Th1, Th2, FINSEQ_1:def 3, FINSEQ_1:def 13;
then E73:
(Sgm (Seg (k + 1))) . j in Seg (k + 1)
by Th3, FUNCT_1:def 5;
then
(Sgm (Seg (k + 1))) . j in (Seg (k + 1)) \ {(k + 1)}
by Th5, E40, XBOOLE_0:def 4;
then E74:
(Sgm (Seg (k + 1))) . j in Seg k
by FINSEQ_1:12;
then reconsider n =
(Sgm (Seg (k + 1))) . j as
Element of
NAT ;
E75:
Sgm (Seg (k + 1)) is
one-to-one
by ;
(Sgm (Seg (k + 1))) | (Seg k) = Sgm (Seg k)
by ;
then E76:
(Sgm (Seg (k + 1))) . n =
(Sgm (Seg k)) . n
by Th13, FUNCT_1:72
.=
n
by , Th13, FUNCT_1:35
;
(
n <= k &
k < k + 1 )
by Th13, FINSEQ_1:3, XREAL_1:31;
hence
contradiction
by Th5, Th9, Th11, E42, E44, E45, FUNCT_1:def 8;
end;
hence
(Sgm (Seg (k + 1))) . j = (idseq (k + 1)) . j
by Th9, FUNCT_1:35;
end;
end;
end;
hence
(Sgm (Seg (k + 1))) . j = (idseq (k + 1)) . j
;
end;
hence
Sgm (Seg (k + 1)) = idseq (k + 1)
by Th1, Th2, FINSEQ_2:10;
end;
theorem Th54: :: FINSEQ_3:54
theorem Th55: :: FINSEQ_3:55
theorem Th56: :: FINSEQ_3:56
theorem Th57: :: FINSEQ_3:57
theorem Th58: :: FINSEQ_3:58
theorem Th59: :: FINSEQ_3:59
theorem Th60: :: FINSEQ_3:60
theorem Th61: :: FINSEQ_3:61
theorem Th62: :: FINSEQ_3:62
theorem Th63: :: FINSEQ_3:63
theorem Th64: :: FINSEQ_3:64
:: deftheorem Def1 defines - FINSEQ_3:def 1 :
theorem Th65: :: FINSEQ_3:65
canceled;
theorem Th66: :: FINSEQ_3:66
theorem Th67: :: FINSEQ_3:67
theorem Th68: :: FINSEQ_3:68
theorem Th69: :: FINSEQ_3:69
theorem Th70: :: FINSEQ_3:70
theorem Th71: :: FINSEQ_3:71
theorem Th72: :: FINSEQ_3:72
theorem Th73: :: FINSEQ_3:73
theorem Th74: :: FINSEQ_3:74
theorem Th75: :: FINSEQ_3:75
theorem Th76: :: FINSEQ_3:76
theorem Th77: :: FINSEQ_3:77
theorem Th78: :: FINSEQ_3:78
theorem Th79: :: FINSEQ_3:79
Lemma169:
for A being set holds {} - A = {}
Lemma170:
for x, A being set holds
( <*x*> - A = <*x*> iff not x in A )
Lemma171:
for x, A being set holds
( <*x*> - A = {} iff x in A )
Lemma172:
for p being FinSequence
for A being set holds (p ^ {} ) - A = (p - A) ^ ({} - A)
Lemma173:
for p being FinSequence
for x, A being set holds (p ^ <*x*>) - A = (p - A) ^ (<*x*> - A)
Lemma181:
for q, p being FinSequence
for A being set holds (p ^ q) - A = (p - A) ^ (q - A)
theorem Th80: :: FINSEQ_3:80
theorem Th81: :: FINSEQ_3:81
theorem Th82: :: FINSEQ_3:82
theorem Th83: :: FINSEQ_3:83
theorem Th84: :: FINSEQ_3:84
theorem Th85: :: FINSEQ_3:85
theorem Th86: :: FINSEQ_3:86
theorem Th87: :: FINSEQ_3:87
theorem Th88: :: FINSEQ_3:88
theorem Th89: :: FINSEQ_3:89
for
x,
y,
A being
set holds
(
<*x,y*> - A = <*x,y*> iff ( not
x in A & not
y in A ) )
theorem Th90: :: FINSEQ_3:90
theorem Th91: :: FINSEQ_3:91
Lemma187:
for p being FinSequence
for A being set st len p = 0 holds
for n being natural number st n in dom p holds
for B being finite set holds
( not B = { k where k is Element of NAT : ( k in dom p & k <= n & p . k in A ) } or p . n in A or (p - A) . (n - (card B)) = p . n )
Lemma188:
for l being natural number st ( for p being FinSequence
for A being set st len p = l holds
for n being natural number st n in dom p holds
for B being finite set holds
( not B = { k where k is Element of NAT : ( k in dom p & k <= n & p . k in A ) } or p . n in A or (p - A) . (n - (card B)) = p . n ) ) holds
for p being FinSequence
for A being set st len p = l + 1 holds
for n being natural number st n in dom p holds
for C being finite set holds
( not C = { m where m is Element of NAT : ( m in dom p & m <= n & p . m in A ) } or p . n in A or (p - A) . (n - (card C)) = p . n )
Lemma191:
for l being natural number
for p being FinSequence
for A being set st len p = l holds
for n being natural number st n in dom p holds
for B being finite set holds
( not B = { k where k is Element of NAT : ( k in dom p & k <= n & p . k in A ) } or p . n in A or (p - A) . (n - (card B)) = p . n )
theorem Th92: :: FINSEQ_3:92
theorem Th93: :: FINSEQ_3:93
theorem Th94: :: FINSEQ_3:94
theorem Th95: :: FINSEQ_3:95
theorem Th96: :: FINSEQ_3:96
theorem Th97: :: FINSEQ_3:97
theorem Th98: :: FINSEQ_3:98
theorem Th99: :: FINSEQ_3:99
theorem Th100: :: FINSEQ_3:100
theorem Th101: :: FINSEQ_3:101
canceled;
theorem Th102: :: FINSEQ_3:102
theorem Th103: :: FINSEQ_3:103
theorem Th104: :: FINSEQ_3:104
theorem Th105: :: FINSEQ_3:105
theorem Th106: :: FINSEQ_3:106
theorem Th107: :: FINSEQ_3:107
theorem Th108: :: FINSEQ_3:108
theorem Th109: :: FINSEQ_3:109
theorem Th110: :: FINSEQ_3:110
theorem Th111: :: FINSEQ_3:111
theorem Th112: :: FINSEQ_3:112
:: deftheorem Def2 defines Del FINSEQ_3:def 2 :
theorem Th113: :: FINSEQ_3:113
theorem Th114: :: FINSEQ_3:114
theorem Th115: :: FINSEQ_3:115
theorem Th116: :: FINSEQ_3:116
theorem Th117: :: FINSEQ_3:117
theorem Th118: :: FINSEQ_3:118
theorem Th119: :: FINSEQ_3:119
theorem Th120: :: FINSEQ_3:120
theorem Th121: :: FINSEQ_3:121
theorem Th122: :: FINSEQ_3:122
theorem Th123: :: FINSEQ_3:123
theorem Th124: :: FINSEQ_3:124