:: Two Programs for {\bf SCM}. Part I - Preliminaries
:: by Grzegorz Bancerek and Piotr Rudnicki
::
:: Copyright (c) 1993-2011 Association of Mizar Users

begin

definition
let X1, X2 be non empty set ;
let Y1 be non empty Subset of X1;
let Y2 be non empty Subset of X2;
let x be Element of [:Y1,Y2:];
:: original: 1
redefine func x 1 -> Element of Y1;
coherence
x 1 is Element of Y1
proof end;
:: original: 2
redefine func x 2 -> Element of Y2;
coherence
x 2 is Element of Y2
proof end;
end;

definition
let n be Nat;
func Fib n -> Element of NAT means :Def1: :: PRE_FF:def 1
ex fib being Function of NAT, st
( it = (fib . n) 1 & fib . 0 = [0,1] & ( for n being Nat holds fib . (n + 1) = [((fib . n) 2),(((fib . n) 1) + ((fib . n) 2))] ) );
existence
ex b1 being Element of NAT ex fib being Function of NAT, st
( b1 = (fib . n) 1 & fib . 0 = [0,1] & ( for n being Nat holds fib . (n + 1) = [((fib . n) 2),(((fib . n) 1) + ((fib . n) 2))] ) )
proof end;
uniqueness
for b1, b2 being Element of NAT st ex fib being Function of NAT, st
( b1 = (fib . n) 1 & fib . 0 = [0,1] & ( for n being Nat holds fib . (n + 1) = [((fib . n) 2),(((fib . n) 1) + ((fib . n) 2))] ) ) & ex fib being Function of NAT, st
( b2 = (fib . n) 1 & fib . 0 = [0,1] & ( for n being Nat holds fib . (n + 1) = [((fib . n) 2),(((fib . n) 1) + ((fib . n) 2))] ) ) holds
b1 = b2
proof end;
end;

:: deftheorem Def1 defines Fib PRE_FF:def 1 :
for n being Nat
for b2 being Element of NAT holds
( b2 = Fib n iff ex fib being Function of NAT, st
( b2 = (fib . n) 1 & fib . 0 = [0,1] & ( for n being Nat holds fib . (n + 1) = [((fib . n) 2),(((fib . n) 1) + ((fib . n) 2))] ) ) );

theorem :: PRE_FF:1
( Fib 0 = 0 & Fib 1 = 1 & ( for n being Nat holds Fib ((n + 1) + 1) = (Fib n) + (Fib (n + 1)) ) )
proof end;

theorem :: PRE_FF:2
for i being Integer holds i div 1 = i
proof end;

theorem :: PRE_FF:3
for i, j being Integer st j > 0 & i div j = 0 holds
i < j
proof end;

theorem :: PRE_FF:4
for i, j being Integer st 0 <= i & i < j holds
i div j = 0
proof end;

theorem :: PRE_FF:5
for i, j, k being Integer st k > 0 holds
(i div j) div k = i div (j * k)
proof end;

theorem :: PRE_FF:6
for i being Integer holds
( i mod 2 = 0 or i mod 2 = 1 )
proof end;

theorem :: PRE_FF:7
for i being Integer st i is Element of NAT holds
i div 2 is Element of NAT
proof end;

theorem :: PRE_FF:8
canceled;

theorem :: PRE_FF:9
canceled;

theorem Th10: :: PRE_FF:10
for a, b, c being real number st a <= b & c > 1 holds
c to_power a <= c to_power b
proof end;

theorem Th11: :: PRE_FF:11
for r, s being real number st r >= s holds
[\r/] >= [\s/]
proof end;

theorem Th12: :: PRE_FF:12
for a, b, c being real number st a > 1 & b > 0 & c >= b holds
log (a,c) >= log (a,b)
proof end;

theorem Th13: :: PRE_FF:13
for n being Nat st n > 0 holds
[\(log (2,(2 * n)))/] + 1 <> [\(log (2,((2 * n) + 1)))/]
proof end;

theorem Th14: :: PRE_FF:14
for n being Nat st n > 0 holds
[\(log (2,(2 * n)))/] + 1 >= [\(log (2,((2 * n) + 1)))/]
proof end;

theorem Th15: :: PRE_FF:15
for n being Nat st n > 0 holds
[\(log (2,(2 * n)))/] = [\(log (2,((2 * n) + 1)))/]
proof end;

theorem :: PRE_FF:16
for n being Nat st n > 0 holds
[\(log (2,n))/] + 1 = [\(log (2,((2 * n) + 1)))/]
proof end;

definition
let f be Function of NAT,();
let n be Element of NAT ;
:: original: .
redefine func f . n -> FinSequence of NAT ;
coherence
f . n is FinSequence of NAT
proof end;
end;

defpred S1[ Nat, FinSequence of NAT , set ] means ( ( for k being Nat st $1 + 2 = 2 * k holds$3 = $2 ^ <*($2 /. k)*> ) & ( for k being Nat st $1 + 2 = (2 * k) + 1 holds$3 = $2 ^ <*(($2 /. k) + ($2 /. (k + 1)))*> ) ); Lm1: for n being Element of NAT for x being Element of NAT * ex y being Element of NAT * st S1[n,x,y] proof end; defpred S2[ Nat, FinSequence of NAT , set ] means ( ( for k being Element of NAT st$1 + 2 = 2 * k holds
$3 =$2 ^ <*($2 /. k)*> ) & ( for k being Element of NAT st$1 + 2 = (2 * k) + 1 holds
$3 =$2 ^ <*(($2 /. k) + ($2 /. (k + 1)))*> ) );

Lm2: for n being Nat
for x, y1, y2 being Element of NAT * st S2[n,x,y1] & S2[n,x,y2] holds
y1 = y2
proof end;

reconsider single1 = <*1*> as Element of NAT * by FINSEQ_1:def 11;

consider fusc being Function of NAT,() such that
Lm3: fusc . 0 = single1 and
Lm4: for n being Element of NAT holds S1[n,fusc . n,fusc . (n + 1)] from
Lm5: for n being Nat holds S1[n,fusc . n,fusc . (n + 1)]
proof end;

definition
let n be Nat;
func Fusc n -> Element of NAT means :Def2: :: PRE_FF:def 2
it = 0 if n = 0
otherwise ex l being Element of NAT ex fusc being Function of NAT,() st
( l + 1 = n & it = (fusc . l) /. n & fusc . 0 = <*1*> & ( for n being Nat holds
( ( for k being Nat st n + 2 = 2 * k holds
fusc . (n + 1) = (fusc . n) ^ <*((fusc . n) /. k)*> ) & ( for k being Nat st n + 2 = (2 * k) + 1 holds
fusc . (n + 1) = (fusc . n) ^ <*(((fusc . n) /. k) + ((fusc . n) /. (k + 1)))*> ) ) ) );
consistency
for b1 being Element of NAT holds verum
;
existence
( ( n = 0 implies ex b1 being Element of NAT st b1 = 0 ) & ( not n = 0 implies ex b1, l being Element of NAT ex fusc being Function of NAT,() st
( l + 1 = n & b1 = (fusc . l) /. n & fusc . 0 = <*1*> & ( for n being Nat holds
( ( for k being Nat st n + 2 = 2 * k holds
fusc . (n + 1) = (fusc . n) ^ <*((fusc . n) /. k)*> ) & ( for k being Nat st n + 2 = (2 * k) + 1 holds
fusc . (n + 1) = (fusc . n) ^ <*(((fusc . n) /. k) + ((fusc . n) /. (k + 1)))*> ) ) ) ) ) )
proof end;
uniqueness
for b1, b2 being Element of NAT holds
( ( n = 0 & b1 = 0 & b2 = 0 implies b1 = b2 ) & ( not n = 0 & ex l being Element of NAT ex fusc being Function of NAT,() st
( l + 1 = n & b1 = (fusc . l) /. n & fusc . 0 = <*1*> & ( for n being Nat holds
( ( for k being Nat st n + 2 = 2 * k holds
fusc . (n + 1) = (fusc . n) ^ <*((fusc . n) /. k)*> ) & ( for k being Nat st n + 2 = (2 * k) + 1 holds
fusc . (n + 1) = (fusc . n) ^ <*(((fusc . n) /. k) + ((fusc . n) /. (k + 1)))*> ) ) ) ) & ex l being Element of NAT ex fusc being Function of NAT,() st
( l + 1 = n & b2 = (fusc . l) /. n & fusc . 0 = <*1*> & ( for n being Nat holds
( ( for k being Nat st n + 2 = 2 * k holds
fusc . (n + 1) = (fusc . n) ^ <*((fusc . n) /. k)*> ) & ( for k being Nat st n + 2 = (2 * k) + 1 holds
fusc . (n + 1) = (fusc . n) ^ <*(((fusc . n) /. k) + ((fusc . n) /. (k + 1)))*> ) ) ) ) implies b1 = b2 ) )
proof end;
end;

:: deftheorem Def2 defines Fusc PRE_FF:def 2 :
for n being Nat
for b2 being Element of NAT holds
( ( n = 0 implies ( b2 = Fusc n iff b2 = 0 ) ) & ( not n = 0 implies ( b2 = Fusc n iff ex l being Element of NAT ex fusc being Function of NAT,() st
( l + 1 = n & b2 = (fusc . l) /. n & fusc . 0 = <*1*> & ( for n being Nat holds
( ( for k being Nat st n + 2 = 2 * k holds
fusc . (n + 1) = (fusc . n) ^ <*((fusc . n) /. k)*> ) & ( for k being Nat st n + 2 = (2 * k) + 1 holds
fusc . (n + 1) = (fusc . n) ^ <*(((fusc . n) /. k) + ((fusc . n) /. (k + 1)))*> ) ) ) ) ) ) );

theorem Th17: :: PRE_FF:17
( Fusc 0 = 0 & Fusc 1 = 1 & ( for n being Nat holds
( Fusc (2 * n) = Fusc n & Fusc ((2 * n) + 1) = (Fusc n) + (Fusc (n + 1)) ) ) )
proof end;

theorem :: PRE_FF:18
for nn, nn9 being Nat st nn <> 0 & nn = 2 * nn9 holds
nn9 < nn
proof end;

theorem :: PRE_FF:19
for nn, nn9 being Nat st nn = (2 * nn9) + 1 holds
nn9 < nn
proof end;

theorem :: PRE_FF:20
for A, B being Nat holds B = (A * ()) + (B * (Fusc (0 + 1))) by Th17;

theorem :: PRE_FF:21
for nn, nn9, A, B, N being Nat st nn = (2 * nn9) + 1 & Fusc N = (A * (Fusc nn)) + (B * (Fusc (nn + 1))) holds
Fusc N = (A * (Fusc nn9)) + ((B + A) * (Fusc (nn9 + 1)))
proof end;

theorem :: PRE_FF:22
for nn, nn9, A, B, N being Nat st nn = 2 * nn9 & Fusc N = (A * (Fusc nn)) + (B * (Fusc (nn + 1))) holds
Fusc N = ((A + B) * (Fusc nn9)) + (B * (Fusc (nn9 + 1)))
proof end;