set a = intloc 0 ;
let s be State of ; :: thesis: for I being good Program of st I is_halting_on Initialize s & I is_closed_on Initialize s holds
( (IExec I,s) . (intloc 0 ) = 1 & ( for k being Element of NAT holds (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),k) . (intloc 0 ) = 1 ) )
set A = NAT ;
let I be good Program of ; :: thesis: ( I is_halting_on Initialize s & I is_closed_on Initialize s implies ( (IExec I,s) . (intloc 0 ) = 1 & ( for k being Element of NAT holds (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),k) . (intloc 0 ) = 1 ) ) )
set s0 = Initialize s;
set s1 = (Initialize s) +* (I +* (Start-At (insloc 0 )));
defpred S₁[ Element of NAT ] means for n being Element of NAT st n <= $1 holds
(Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),n) . (intloc 0 ) = (Initialize s) . (intloc 0 );
assume I is_halting_on Initialize s ; :: thesis: ( not I is_closed_on Initialize s or ( (IExec I,s) . (intloc 0 ) = 1 & ( for k being Element of NAT holds (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),k) . (intloc 0 ) = 1 ) ) )
then A1: ProgramPart ((Initialize s) +* (I +* (Start-At (insloc 0 )))) halts_on (Initialize s) +* (I +* (Start-At (insloc 0 ))) by SCMFSA7B:def 8;
A2: S₁[ 0 ]

proof

let n be Element of NAT ; :: thesis: ( n <= 0 implies (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),n) . (intloc 0 ) = (Initialize s) . (intloc 0 ) )
A3: for i being Element of NAT st i < 0 holds
IC (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),i) in dom I ;
assume n <= 0 ; :: thesis: (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),n) . (intloc 0 ) = (Initialize s) . (intloc 0 )
hence (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),n) . (intloc 0 ) = (Initialize s) . (intloc 0 ) by A3, Th95; :: thesis: verum

end;

assume A4: I is_closed_on Initialize s ; :: thesis: ( (IExec I,s) . (intloc 0 ) = 1 & ( for k being Element of NAT holds (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),k) . (intloc 0 ) = 1 ) )
A5: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume S₁[k] ; :: thesis: S₁[k + 1]
let n be Element of NAT ; :: thesis: ( n <= k + 1 implies (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),n) . (intloc 0 ) = (Initialize s) . (intloc 0 ) )
assume A6: n <= k + 1 ; :: thesis: (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),n) . (intloc 0 ) = (Initialize s) . (intloc 0 )
for i being Element of NAT st i < k + 1 holds
IC (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),i) in dom I by A4, SCMFSA7B:def 7;
hence (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),n) . (intloc 0 ) = (Initialize s) . (intloc 0 ) by A6, Th95; :: thesis: verum

end;

A7: for k being Element of NAT holds S₁[k] from NAT_1:sch 1(A2, A5);

A8: now

let k be Element of NAT ; :: thesis: (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),k) . (intloc 0 ) = 1
thus (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),k) . (intloc 0 ) = (Initialize s) . (intloc 0 ) by A7
.= 1 by SCMFSA6C:3 ; :: thesis: verum

end;

now

assume A9: intloc 0 in NAT ; :: thesis: contradiction
then reconsider a = intloc 0 as Instruction-Location of SCM+FSA by AMI_1:def 4;
a in NAT by A9;
hence contradiction by SCMFSA_2:84; :: thesis: verum

end;

then not intloc 0 in (dom s) /\ NAT by XBOOLE_0:def 4;
then A10: not intloc 0 in dom (s | NAT ) by RELAT_1:90;
s +* (Initialized I) = (Initialize s) +* (I +* (Start-At (insloc 0 ))) by SCMFSA8A:13;
hence (IExec I,s) . (intloc 0 ) = ((Result ((Initialize s) +* (I +* (Start-At (insloc 0 ))))) +* (s | NAT )) . (intloc 0 ) by SCMFSA6B:def 1
.= (Result ((Initialize s) +* (I +* (Start-At (insloc 0 ))))) . (intloc 0 ) by A10, FUNCT_4:12
.= (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),(LifeSpan ((Initialize s) +* (I +* (Start-At (insloc 0 )))))) . (intloc 0 ) by A1, AMI_1:122
.= 1 by A8 ;
:: thesis: for k being Element of NAT holds (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),k) . (intloc 0 ) = 1
thus for k being Element of NAT holds (Computation ((Initialize s) +* (I +* (Start-At (insloc 0 )))),k) . (intloc 0 ) = 1 by A8; :: thesis: verum