TSTP Solution File: DAT100_1 by SPASS+T---2.2.22

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : SPASS+T---2.2.22
% Problem  : DAT100_1 : TPTP v8.1.0. Released v6.1.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : spasst-tptp-script %s %d

% Computer : n021.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  : 600s
% DateTime : Sat Jul 16 01:32:13 EDT 2022

% Result   : Theorem 0.45s 1.06s
% Output   : Refutation 0.45s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.12  % Problem  : DAT100_1 : TPTP v8.1.0. Released v6.1.0.
% 0.12/0.13  % Command  : spasst-tptp-script %s %d
% 0.12/0.34  % Computer : n021.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit : 300
% 0.12/0.34  % WCLimit  : 600
% 0.12/0.34  % DateTime : Fri Jul  1 17:46:29 EDT 2022
% 0.12/0.34  % CPUTime  : 
% 0.19/0.48  % Using integer theory
% 0.45/1.06  
% 0.45/1.06  
% 0.45/1.06  % SZS status Theorem for /tmp/SPASST_7065_n021.cluster.edu
% 0.45/1.06  
% 0.45/1.06  SPASS V 2.2.22  in combination with yices.
% 0.45/1.06  SPASS beiseite: Proof found by SMT.
% 0.45/1.06  Problem: /tmp/SPASST_7065_n021.cluster.edu 
% 0.45/1.06  SPASS derived 79 clauses, backtracked 0 clauses and kept 76 clauses.
% 0.45/1.06  SPASS backtracked 1 times (1 times due to theory inconsistency).
% 0.45/1.06  SPASS allocated 6362 KBytes.
% 0.45/1.06  SPASS spent	0:00:00.02 on the problem.
% 0.45/1.06  		0:00:00.00 for the input.
% 0.45/1.06  		0:00:00.01 for the FLOTTER CNF translation.
% 0.45/1.06  		0:00:00.00 for inferences.
% 0.45/1.06  		0:00:00.00 for the backtracking.
% 0.45/1.06  		0:00:00.01 for the reduction.
% 0.45/1.06  		0:00:00.00 for interacting with the SMT procedure.
% 0.45/1.06  		
% 0.45/1.06  
% 0.45/1.06  % SZS output start CNFRefutation for /tmp/SPASST_7065_n021.cluster.edu
% 0.45/1.06  
% 0.45/1.06  % Here is a proof with depth 0, length 11 :
% 0.45/1.06  2[0:Inp] ||  -> list(nil)*.
% 0.45/1.06  5[0:Inp] ||  -> inRange(U,skf2(U,V))*.
% 0.45/1.06  7[0:Inp] || list(U) -> list(cons(V,U))*.
% 0.45/1.06  8[0:Inp] || list(U) equal(cons(V,U),nil)** -> .
% 0.45/1.06  9[0:Inp] || list(U) -> equal(tail(cons(V,U)),U)**.
% 0.45/1.06  10[0:Inp] || list(U) -> equal(head(cons(V,U)),V)**.
% 0.45/1.06  11[0:Inp] ||  -> inRange(4,cons(1,cons(5,cons(2,nil))))*.
% 0.45/1.06  14[0:Inp] || list(U) inRange(V,U) -> list(skf2(V,U))* equal(U,nil).
% 0.45/1.06  15[0:Inp] || list(U) inRange(V,U) -> equal(U,nil) less(skf3(V,U),V)*.
% 0.45/1.06  16[0:Inp] || list(U) inRange(V,U) -> equal(U,nil) equal(cons(skf3(V,U),skf2(V,U)),U)**.
% 0.45/1.06  156(e)[0:ThR:16,15,14,11,10,9,8,7,5,2] ||  -> .
% 0.45/1.06  
% 0.45/1.06  % SZS output end CNFRefutation for /tmp/SPASST_7065_n021.cluster.edu
% 0.45/1.06  
% 0.45/1.06  Formulae used in the proof : fof_head_type fof_inRange fof_nil_type fof_cons_type fof_tail_type fof_l2 fof_l4 fof_l1 fof_l3
% 0.45/1.06  
% 0.45/1.06  SPASS+T ended
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