TSTP Solution File: NUM530+1 by SOS---2.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : SOS---2.0
% Problem  : NUM530+1 : TPTP v8.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : sos-script %s

% Computer : n005.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 : Mon Jul 18 14:19:11 EDT 2022

% Result   : Theorem 0.20s 0.38s
% Output   : Refutation 0.20s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12  % Problem  : NUM530+1 : TPTP v8.1.0. Released v4.0.0.
% 0.03/0.13  % Command  : sos-script %s
% 0.12/0.34  % Computer : n005.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 : Tue Jul  5 18:41:22 EDT 2022
% 0.12/0.34  % CPUTime  : 
% 0.20/0.38  
% 0.20/0.38  -------- PROOF -------- 
% 0.20/0.38  % SZS status Unsatisfiable
% 0.20/0.38  % SZS output start Refutation
% 0.20/0.38  ----- Otter 3.2, August 2001 -----
% 0.20/0.38  The process was started by sandbox on n005.cluster.edu,
% 0.20/0.38  Tue Jul  5 18:41:22 2022
% 0.20/0.38  The command was "./sos".  The process ID is 24204.
% 0.20/0.38  
% 0.20/0.38  set(prolog_style_variables).
% 0.20/0.38  set(auto).
% 0.20/0.38     dependent: set(auto1).
% 0.20/0.38     dependent: set(process_input).
% 0.20/0.38     dependent: clear(print_kept).
% 0.20/0.38     dependent: clear(print_new_demod).
% 0.20/0.38     dependent: clear(print_back_demod).
% 0.20/0.38     dependent: clear(print_back_sub).
% 0.20/0.38     dependent: set(control_memory).
% 0.20/0.38     dependent: assign(max_mem, 12000).
% 0.20/0.38     dependent: assign(pick_given_ratio, 4).
% 0.20/0.38     dependent: assign(stats_level, 1).
% 0.20/0.38     dependent: assign(pick_semantic_ratio, 3).
% 0.20/0.38     dependent: assign(sos_limit, 5000).
% 0.20/0.38     dependent: assign(max_weight, 60).
% 0.20/0.38  clear(print_given).
% 0.20/0.38  
% 0.20/0.38  formula_list(usable).
% 0.20/0.38  
% 0.20/0.38  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=9.
% 0.20/0.38  
% 0.20/0.38  This ia a non-Horn set with equality.  The strategy will be
% 0.20/0.38  Knuth-Bendix, ordered hyper_res, ur_res, factoring, and
% 0.20/0.38  unit deletion, with positive clauses in sos and nonpositive
% 0.20/0.38  clauses in usable.
% 0.20/0.38  
% 0.20/0.38     dependent: set(knuth_bendix).
% 0.20/0.38     dependent: set(para_from).
% 0.20/0.38     dependent: set(para_into).
% 0.20/0.38     dependent: clear(para_from_right).
% 0.20/0.38     dependent: clear(para_into_right).
% 0.20/0.38     dependent: set(para_from_vars).
% 0.20/0.38     dependent: set(eq_units_both_ways).
% 0.20/0.38     dependent: set(dynamic_demod_all).
% 0.20/0.38     dependent: set(dynamic_demod).
% 0.20/0.38     dependent: set(order_eq).
% 0.20/0.38     dependent: set(back_demod).
% 0.20/0.38     dependent: set(lrpo).
% 0.20/0.38     dependent: set(hyper_res).
% 0.20/0.38     dependent: set(unit_deletion).
% 0.20/0.38     dependent: set(factor).
% 0.20/0.38  
% 0.20/0.38  ------------> process usable:
% 0.20/0.38    Following clause subsumed by 21 during input processing: 0 [] {-} -aNaturalNumber0(A)| -aNaturalNumber0(B)|A=B| -sdtlseqdt0(A,B)| -aNaturalNumber0(C)|sdtpldt0(C,A)!=sdtpldt0(C,B).
% 0.20/0.38    Following clause subsumed by 22 during input processing: 0 [] {-} -aNaturalNumber0(A)| -aNaturalNumber0(B)|A=B| -sdtlseqdt0(A,B)| -aNaturalNumber0(C)|sdtpldt0(A,C)!=sdtpldt0(B,C).
% 0.20/0.38    Following clause subsumed by 23 during input processing: 0 [] {-} -aNaturalNumber0(A)| -aNaturalNumber0(B)| -aNaturalNumber0(C)|A=sz00|B=C| -sdtlseqdt0(B,C)|sdtasdt0(A,B)!=sdtasdt0(A,C).
% 0.20/0.38    Following clause subsumed by 24 during input processing: 0 [] {-} -aNaturalNumber0(A)| -aNaturalNumber0(B)| -aNaturalNumber0(C)|A=sz00|B=C| -sdtlseqdt0(B,C)|sdtasdt0(B,A)!=sdtasdt0(C,A).
% 0.20/0.38  
% 0.20/0.38  ------------> process sos:
% 0.20/0.38  
% 0.20/0.38  ----> UNIT CONFLICT at   0.02 sec ----> 181 [binary,180.1,77.1] {-} $F.
% 0.20/0.38  
% 0.20/0.38  Length of proof is 0.  Level of proof is 0.
% 0.20/0.38  
% 0.20/0.38  ---------------- PROOF ----------------
% 0.20/0.38  % SZS status Theorem
% 0.20/0.38  % SZS output start Refutation
% 0.20/0.38  
% 0.20/0.38  77 [] {-} -isPrime0(xp).
% 0.20/0.38  180 [] {-} isPrime0(xp).
% 0.20/0.38  181 [binary,180.1,77.1] {-} $F.
% 0.20/0.38  
% 0.20/0.38  % SZS output end Refutation
% 0.20/0.38  ------------ end of proof -------------
% 0.20/0.38  
% 0.20/0.38  
% 0.20/0.38  Search stopped by max_proofs option.
% 0.20/0.38  
% 0.20/0.38  
% 0.20/0.38  Search stopped by max_proofs option.
% 0.20/0.38  
% 0.20/0.38  ============ end of search ============
% 0.20/0.38  
% 0.20/0.38  That finishes the proof of the theorem.
% 0.20/0.38  
% 0.20/0.38  Process 24204 finished Tue Jul  5 18:41:22 2022
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