%------------------------------------------------------------------------------ % File : cocATP---0.2.0 % Problem : SEV286^5 : TPTP v6.1.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p % Computer : n117.star.cs.uiowa.edu % Model : x86_64 x86_64 % CPU : Intel(R) Xeon(R) CPU E5-2609 0 2.40GHz % Memory : 32286.75MB % OS : Linux 2.6.32-431.20.3.el6.x86_64 % CPULimit : 300s % DateTime : Thu Jul 17 13:33:59 EDT 2014 % Result : Theorem 0.47s % Output : Proof 0.47s % Verified : % SZS Type : None (Parsing solution fails) % Syntax : Number of formulae : 0 % Comments : %------------------------------------------------------------------------------ %----ERROR: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % % Problem : SEV286^5 : TPTP v6.1.0. Released v4.0.0. % % Command : python CASC.py /export/starexec/sandbox/benchmark/theBenchmark.p % % Computer : n117.star.cs.uiowa.edu % % Model : x86_64 x86_64 % % CPU : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz % % Memory : 32286.75MB % % OS : Linux 2.6.32-431.20.3.el6.x86_64 % % CPULimit : 300 % % DateTime : Thu Jul 17 08:44:11 CDT 2014 % % CPUTime : 0.47 % Python 2.7.5 % Using paths ['/home/cristobal/cocATP/CASC/TPTP/', '/export/starexec/sandbox/benchmark/', '/export/starexec/sandbox/benchmark/'] % FOF formula (<kernel.Constant object at 0x1454098>, <kernel.Type object at 0x1453878>) of role type named a_type % Using role type % Declaring a:Type % FOF formula (<kernel.Constant object at 0x127c908>, <kernel.Type object at 0x1453368>) of role type named b_type % Using role type % Declaring b:Type % FOF formula (forall (F:(a->b)) (A:a) (B:a), ((((eq a) A) B)->(((eq b) (F A)) (F B)))) of role conjecture named cEC_eq_ % Conjecture to prove = (forall (F:(a->b)) (A:a) (B:a), ((((eq a) A) B)->(((eq b) (F A)) (F B)))):Prop % Parameter a_DUMMY:a. % Parameter b_DUMMY:b. % We need to prove ['(forall (F:(a->b)) (A:a) (B:a), ((((eq a) A) B)->(((eq b) (F A)) (F B))))'] % Parameter a:Type. % Parameter b:Type. % Trying to prove (forall (F:(a->b)) (A:a) (B:a), ((((eq a) A) B)->(((eq b) (F A)) (F B)))) % Found eq_substitution00000:=(eq_substitution0000 F):((((eq a) A) B)->(((eq b) (F A)) (F B))) % Found (eq_substitution0000 F) as proof of ((((eq a) A) B)->(((eq b) (F A)) (F B))) % Found ((eq_substitution000 B) F) as proof of ((((eq a) A) B)->(((eq b) (F A)) (F B))) % Found (((eq_substitution00 A) B) F) as proof of ((((eq a) A) B)->(((eq b) (F A)) (F B))) % Found ((((eq_substitution0 b) A) B) F) as proof of ((((eq a) A) B)->(((eq b) (F A)) (F B))) % Found (((((eq_substitution a) b) A) B) F) as proof of ((((eq a) A) B)->(((eq b) (F A)) (F B))) % Found (fun (B:a)=> (((((eq_substitution a) b) A) B) F)) as proof of ((((eq a) A) B)->(((eq b) (F A)) (F B))) % Found (fun (A:a) (B:a)=> (((((eq_substitution a) b) A) B) F)) as proof of (forall (B:a), ((((eq a) A) B)->(((eq b) (F A)) (F B)))) % Found (fun (F:(a->b)) (A:a) (B:a)=> (((((eq_substitution a) b) A) B) F)) as proof of (forall (A:a) (B:a), ((((eq a) A) B)->(((eq b) (F A)) (F B)))) % Found (fun (F:(a->b)) (A:a) (B:a)=> (((((eq_substitution a) b) A) B) F)) as proof of (forall (F:(a->b)) (A:a) (B:a), ((((eq a) A) B)->(((eq b) (F A)) (F B)))) % Got proof (fun (F:(a->b)) (A:a) (B:a)=> (((((eq_substitution a) b) A) B) F)) % Time elapsed = 0.157515s % node=9 cost=-454.000000 depth=8 % :::::::::::::::::::::: % % SZS status Theorem for /export/starexec/sandbox/benchmark/theBenchmark.p % % SZS output start Proof for /export/starexec/sandbox/benchmark/theBenchmark.p % (fun (F:(a->b)) (A:a) (B:a)=> (((((eq_substitution a) b) A) B) F)) % % SZS output end Proof for /export/starexec/sandbox/benchmark/theBenchmark.p % EOF %------------------------------------------------------------------------------