let z be Nat; for N being with_zero set
for l being Nat st l = z holds
SUCC (l,(STC N)) = {z,(z + 1)}
let N be with_zero set ; for l being Nat st l = z holds
SUCC (l,(STC N)) = {z,(z + 1)}
let l be Nat; ( l = z implies SUCC (l,(STC N)) = {z,(z + 1)} )
assume A1:
l = z
; SUCC (l,(STC N)) = {z,(z + 1)}
set M = STC N;
set K = { ((NIC (i,l)) \ (JUMP i)) where i is Element of the InstructionsF of (STC N) : verum } ;
now for y being object holds
( ( y in { ((NIC (i,l)) \ (JUMP i)) where i is Element of the InstructionsF of (STC N) : verum } implies y in {{z},{(z + 1)}} ) & ( y in {{z},{(z + 1)}} implies y in { ((NIC (i,l)) \ (JUMP i)) where i is Element of the InstructionsF of (STC N) : verum } ) )let y be
object ;
( ( y in { ((NIC (i,l)) \ (JUMP i)) where i is Element of the InstructionsF of (STC N) : verum } implies y in {{z},{(z + 1)}} ) & ( y in {{z},{(z + 1)}} implies b1 in { ((NIC (b2,l)) \ (JUMP b2)) where i is Element of the InstructionsF of (STC N) : verum } ) )assume A5:
y in {{z},{(z + 1)}}
;
b1 in { ((NIC (b2,l)) \ (JUMP b2)) where i is Element of the InstructionsF of (STC N) : verum } per cases
( y = {z} or y = {(z + 1)} )
by A5, TARSKI:def 2;
suppose A7:
y = {(z + 1)}
;
b1 in { ((NIC (b2,l)) \ (JUMP b2)) where i is Element of the InstructionsF of (STC N) : verum } set i =
[1,0,0];
[1,0,0] in {[1,{},{}],[0,{},{}]}
by TARSKI:def 2;
then reconsider i =
[1,0,0] as
Instruction of
(STC N) by Def7;
(
JUMP i = {} &
InsCode i = 1 )
by Lm5;
then
(NIC (i,l)) \ (JUMP i) = y
by A1, A7, Lm4;
hence
y in { ((NIC (i,l)) \ (JUMP i)) where i is Element of the InstructionsF of (STC N) : verum }
;
verum end; end; end;
then
{ ((NIC (i,l)) \ (JUMP i)) where i is Element of the InstructionsF of (STC N) : verum } = {{z},{(z + 1)}}
by TARSKI:2;
hence
SUCC (l,(STC N)) = {z,(z + 1)}
by ZFMISC_1:26; verum