TSTP Solution File: SEV406^5 by Vampire---4.9
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%------------------------------------------------------------------------------
% File : Vampire---4.9
% Problem : SEV406^5 : TPTP v8.2.0. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : run_vampire %s %d THM
% Computer : n020.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 300s
% DateTime : Mon Jun 24 16:03:45 EDT 2024
% Result : Theorem 0.20s 0.38s
% Output : Refutation 0.20s
% Verified :
% SZS Type : Refutation
% Derivation depth : 7
% Number of leaves : 1
% Syntax : Number of formulae : 8 ( 5 unt; 0 typ; 0 def)
% Number of atoms : 41 ( 4 equ; 0 cnn)
% Maximal formula atoms : 6 ( 5 avg)
% Number of connectives : 50 ( 5 ~; 10 |; 0 &; 30 @)
% ( 0 <=>; 5 =>; 0 <=; 0 <~>)
% Maximal formula depth : 7 ( 3 avg)
% Number of types : 2 ( 0 usr)
% Number of type conns : 0 ( 0 >; 0 *; 0 +; 0 <<)
% Number of symbols : 6 ( 3 usr; 2 con; 0-2 aty)
% Number of variables : 10 ( 10 ^ 0 !; 0 ?; 10 :)
% Comments :
%------------------------------------------------------------------------------
thf(func_def_0,type,
cA: $i > $o ).
thf(func_def_1,type,
cB: $i > $o ).
thf(func_def_2,type,
cP: ( $i > $o ) > $o ).
thf(f9,plain,
$false,
inference(trivial_inequality_removal,[],[f8]) ).
thf(f8,plain,
$true = $false,
inference(boolean_simplification,[],[f7]) ).
thf(f7,plain,
( $true = ~ $true ),
inference(boolean_simplification,[],[f6]) ).
thf(f6,plain,
( ( ~ ( ( cP
@ ^ [Y0: $i] :
( ( cB @ Y0 )
| ( cA @ Y0 ) ) )
=> ( cP
@ ^ [Y0: $i] :
( ( cB @ Y0 )
| ( cA @ Y0 ) ) ) ) )
= $true ),
inference(cnf_transformation,[],[f5]) ).
thf(f5,plain,
( ( ~ ( ( cP
@ ^ [Y0: $i] :
( ( cB @ Y0 )
| ( cA @ Y0 ) ) )
=> ( cP
@ ^ [Y0: $i] :
( ( cB @ Y0 )
| ( cA @ Y0 ) ) ) ) )
= $true ),
inference(fool_elimination,[],[f4]) ).
thf(f4,plain,
~ ( ( cP
@ ^ [X0: $i] :
( ( cA @ X0 )
| ( cB @ X0 ) ) )
=> ( cP
@ ^ [X1: $i] :
( ( cA @ X1 )
| ( cB @ X1 ) ) ) ),
inference(rectify,[],[f2]) ).
thf(f2,negated_conjecture,
~ ( ( cP
@ ^ [X0: $i] :
( ( cA @ X0 )
| ( cB @ X0 ) ) )
=> ( cP
@ ^ [X0: $i] :
( ( cA @ X0 )
| ( cB @ X0 ) ) ) ),
inference(negated_conjecture,[],[f1]) ).
thf(f1,conjecture,
( ( cP
@ ^ [X0: $i] :
( ( cA @ X0 )
| ( cB @ X0 ) ) )
=> ( cP
@ ^ [X0: $i] :
( ( cA @ X0 )
| ( cB @ X0 ) ) ) ),
file('/export/starexec/sandbox2/benchmark/theBenchmark.p',cTRIVEXT2_pme) ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.12 % Problem : SEV406^5 : TPTP v8.2.0. Released v4.0.0.
% 0.12/0.12 % Command : run_vampire %s %d THM
% 0.12/0.33 % Computer : n020.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Fri Jun 21 18:26:24 EDT 2024
% 0.12/0.33 % CPUTime :
% 0.20/0.35 This is a TH0_THM_NEQ_NAR problem
% 0.20/0.35 Running higher-order theorem proving
% 0.20/0.35 Running /export/starexec/sandbox2/solver/bin/vampire_ho --cores 7 --input_syntax tptp --proof tptp --output_axiom_names on --mode portfolio --schedule snake_tptp_hol /export/starexec/sandbox2/benchmark/theBenchmark.p -m 16384 -t 300
% 0.20/0.38 % (22187)lrs+1002_1:128_aac=none:au=on:cnfonf=lazy_not_gen_be_off:sos=all:i=2:si=on:rtra=on_0 on theBenchmark for (2999ds/2Mi)
% 0.20/0.38 % (22183)lrs+1002_1:8_bd=off:fd=off:hud=10:tnu=1:i=183:si=on:rtra=on_0 on theBenchmark for (2999ds/183Mi)
% 0.20/0.38 % (22186)lrs+10_1:1_au=on:inj=on:i=2:si=on:rtra=on_0 on theBenchmark for (2999ds/2Mi)
% 0.20/0.38 % (22184)lrs+10_1:1_c=on:cnfonf=conj_eager:fd=off:fe=off:kws=frequency:spb=intro:i=4:si=on:rtra=on_0 on theBenchmark for (2999ds/4Mi)
% 0.20/0.38 % (22185)dis+1010_1:1_au=on:cbe=off:chr=on:fsr=off:hfsq=on:nm=64:sos=theory:sp=weighted_frequency:i=27:si=on:rtra=on_0 on theBenchmark for (2999ds/27Mi)
% 0.20/0.38 % (22189)lrs+1004_1:128_cond=on:e2e=on:sp=weighted_frequency:i=18:si=on:rtra=on_0 on theBenchmark for (2999ds/18Mi)
% 0.20/0.38 % (22188)lrs+1002_1:1_au=on:bd=off:e2e=on:sd=2:sos=on:ss=axioms:i=275:si=on:rtra=on_0 on theBenchmark for (2999ds/275Mi)
% 0.20/0.38 % (22187)First to succeed.
% 0.20/0.38 % (22186)Refutation not found, incomplete strategy
% 0.20/0.38 % (22186)------------------------------
% 0.20/0.38 % (22186)Version: Vampire 4.8 (commit 11aac991b on 2023-10-04 16:26:07 +0200)
% 0.20/0.38 % (22186)Termination reason: Refutation not found, incomplete strategy
% 0.20/0.38
% 0.20/0.38
% 0.20/0.38 % (22186)Memory used [KB]: 5500
% 0.20/0.38 % (22186)Time elapsed: 0.004 s
% 0.20/0.38 % (22186)Instructions burned: 1 (million)
% 0.20/0.38 % (22186)------------------------------
% 0.20/0.38 % (22186)------------------------------
% 0.20/0.38 % (22185)Refutation not found, incomplete strategy
% 0.20/0.38 % (22185)------------------------------
% 0.20/0.38 % (22185)Version: Vampire 4.8 (commit 11aac991b on 2023-10-04 16:26:07 +0200)
% 0.20/0.38 % (22185)Termination reason: Refutation not found, incomplete strategy
% 0.20/0.38
% 0.20/0.38
% 0.20/0.38 % (22185)Memory used [KB]: 5500
% 0.20/0.38 % (22185)Time elapsed: 0.003 s
% 0.20/0.38 % (22185)Instructions burned: 1 (million)
% 0.20/0.38 % (22185)------------------------------
% 0.20/0.38 % (22185)------------------------------
% 0.20/0.38 % (22184)Also succeeded, but the first one will report.
% 0.20/0.38 % (22188)Also succeeded, but the first one will report.
% 0.20/0.38 % (22187)Refutation found. Thanks to Tanya!
% 0.20/0.38 % SZS status Theorem for theBenchmark
% 0.20/0.38 % SZS output start Proof for theBenchmark
% See solution above
% 0.20/0.38 % (22187)------------------------------
% 0.20/0.38 % (22187)Version: Vampire 4.8 (commit 11aac991b on 2023-10-04 16:26:07 +0200)
% 0.20/0.38 % (22187)Termination reason: Refutation
% 0.20/0.38
% 0.20/0.38 % (22187)Memory used [KB]: 5373
% 0.20/0.38 % (22187)Time elapsed: 0.004 s
% 0.20/0.38 % (22187)Instructions burned: 1 (million)
% 0.20/0.38 % (22187)------------------------------
% 0.20/0.38 % (22187)------------------------------
% 0.20/0.38 % (22180)Success in time 0.019 s
%------------------------------------------------------------------------------