:: Two Programs for {\bf SCM}. Part I - Preliminaries
:: by Grzegorz Bancerek and Piotr Rudnicki
::
:: Copyright (c) 1993-2021 Association of Mizar Users

:: Definition of fib
registration
let n, k be Nat;
reduce In ([n,k],) to [n,k];
reducibility
In ([n,k],) = [n,k]
proof end;
end;

deffunc H1( set , Element of ) -> Element of = In ([($2 2),(($2 1) + ($2 2))],); definition func Fib -> sequence of means :Def1: :: PRE_FF:def 1 ( it . 0 = [0,1] & ( for n being Nat holds it . (n + 1) = [((it . n) 2),(((it . n) 1) + ((it . n) 2))] ) ); existence ex b1 being sequence of st ( b1 . 0 = [0,1] & ( for n being Nat holds b1 . (n + 1) = [((b1 . n) 2),(((b1 . n) 1) + ((b1 . n) 2))] ) ) proof end; uniqueness for b1, b2 being sequence of st b1 . 0 = [0,1] & ( for n being Nat holds b1 . (n + 1) = [((b1 . n) 2),(((b1 . n) 1) + ((b1 . n) 2))] ) & b2 . 0 = [0,1] & ( for n being Nat holds b2 . (n + 1) = [((b2 . n) 2),(((b2 . n) 1) + ((b2 . n) 2))] ) holds b1 = b2 proof end; end; :: deftheorem Def1 defines Fib PRE_FF:def 1 : for b1 being sequence of holds ( b1 = Fib iff ( b1 . 0 = [0,1] & ( for n being Nat holds b1 . (n + 1) = [((b1 . n) 2),(((b1 . n) 1) + ((b1 . n) 2))] ) ) ); definition let n be Nat; func Fib n -> Element of NAT equals :: PRE_FF:def 2 (Fib . n) 1 ; coherence (Fib . n) 1 is Element of NAT ; end; :: deftheorem defines Fib PRE_FF:def 2 : for n being Nat holds Fib n = (Fib . n) `1 ; 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; :: Fusc function auxiliaries 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 Th8: :: PRE_FF:8 for a, b, c being Real st a <= b & c >= 1 holds c to_power a <= c to_power b proof end; theorem Th9: :: PRE_FF:9 for r, s being Real st r >= s holds [\r/] >= [\s/] proof end; theorem Th10: :: PRE_FF:10 for a, b, c being Real st a > 1 & b > 0 & c >= b holds log (a,c) >= log (a,b) proof end; theorem Th11: :: PRE_FF:11 for n being Nat st n > 0 holds [\(log (2,(2 * n)))/] + 1 <> [\(log (2,((2 * n) + 1)))/] proof end; theorem Th12: :: PRE_FF:12 for n being Nat st n > 0 holds [\(log (2,(2 * n)))/] + 1 >= [\(log (2,((2 * n) + 1)))/] proof end; theorem Th13: :: PRE_FF:13 for n being Nat st n > 0 holds [\(log (2,(2 * n)))/] = [\(log (2,((2 * n) + 1)))/] proof end; theorem :: PRE_FF:14 for n being Nat st n > 0 holds [\(log (2,n))/] + 1 = [\(log (2,((2 * n) + 1)))/] proof 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 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;

definition
let n be Nat;
func Fusc n -> Element of NAT means :Def2: :: PRE_FF:def 3
it = 0 if n = 0
otherwise ex l being Element of NAT ex fusc being sequence of () 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 sequence of () 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 sequence of () 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 sequence of () 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 3 :
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 sequence of () 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 Th15: :: PRE_FF:15
( 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;

:: Auxiliary functions specific for Fib and Fusc of little general interest
theorem :: PRE_FF:16
for n being Nat st n <> 0 holds
n < 2 * n
proof end;

theorem :: PRE_FF:17
for n being Nat holds n < (2 * n) + 1
proof end;

theorem :: PRE_FF:18
for A, B being Nat holds B = (A * ()) + (B * (Fusc 1)) by Th15;

theorem :: PRE_FF:19
for n, A, B, N being Nat st Fusc N = (A * (Fusc ((2 * n) + 1))) + (B * (Fusc (((2 * n) + 1) + 1))) holds
Fusc N = (A * (Fusc n)) + ((B + A) * (Fusc (n + 1)))
proof end;

theorem :: PRE_FF:20
for n, A, B, N being Nat st Fusc N = (A * (Fusc (2 * n))) + (B * (Fusc ((2 * n) + 1))) holds
Fusc N = ((A + B) * (Fusc n)) + (B * (Fusc (n + 1)))
proof end;