TSTP Solution File: SET604+3 by SOS---2.0

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : SOS---2.0
% Problem  : SET604+3 : TPTP v8.1.0. Released v2.2.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : sos-script %s

% Computer : n003.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 : Tue Jul 19 05:19:19 EDT 2022

% Result   : Theorem 0.20s 0.39s
% Output   : Refutation 0.20s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12  % Problem  : SET604+3 : TPTP v8.1.0. Released v2.2.0.
% 0.03/0.12  % Command  : sos-script %s
% 0.14/0.34  % Computer : n003.cluster.edu
% 0.14/0.34  % Model    : x86_64 x86_64
% 0.14/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34  % Memory   : 8042.1875MB
% 0.14/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34  % CPULimit : 300
% 0.14/0.34  % WCLimit  : 600
% 0.14/0.34  % DateTime : Sun Jul 10 14:08:44 EDT 2022
% 0.14/0.34  % CPUTime  : 
% 0.14/0.35  ----- Otter 3.2, August 2001 -----
% 0.14/0.35  The process was started by sandbox on n003.cluster.edu,
% 0.14/0.35  Sun Jul 10 14:08:44 2022
% 0.14/0.35  The command was "./sos".  The process ID is 25898.
% 0.14/0.35  
% 0.14/0.35  set(prolog_style_variables).
% 0.14/0.35  set(auto).
% 0.14/0.35     dependent: set(auto1).
% 0.14/0.35     dependent: set(process_input).
% 0.14/0.35     dependent: clear(print_kept).
% 0.14/0.35     dependent: clear(print_new_demod).
% 0.14/0.35     dependent: clear(print_back_demod).
% 0.14/0.35     dependent: clear(print_back_sub).
% 0.14/0.35     dependent: set(control_memory).
% 0.14/0.35     dependent: assign(max_mem, 12000).
% 0.14/0.35     dependent: assign(pick_given_ratio, 4).
% 0.14/0.35     dependent: assign(stats_level, 1).
% 0.14/0.35     dependent: assign(pick_semantic_ratio, 3).
% 0.14/0.35     dependent: assign(sos_limit, 5000).
% 0.14/0.35     dependent: assign(max_weight, 60).
% 0.14/0.35  clear(print_given).
% 0.14/0.35  
% 0.14/0.35  formula_list(usable).
% 0.14/0.35  
% 0.14/0.35  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 0.14/0.35  
% 0.14/0.35  This ia a non-Horn set with equality.  The strategy will be
% 0.14/0.35  Knuth-Bendix, ordered hyper_res, ur_res, factoring, and
% 0.14/0.35  unit deletion, with positive clauses in sos and nonpositive
% 0.14/0.35  clauses in usable.
% 0.14/0.35  
% 0.14/0.35     dependent: set(knuth_bendix).
% 0.14/0.35     dependent: set(para_from).
% 0.14/0.35     dependent: set(para_into).
% 0.14/0.35     dependent: clear(para_from_right).
% 0.14/0.35     dependent: clear(para_into_right).
% 0.14/0.35     dependent: set(para_from_vars).
% 0.14/0.35     dependent: set(eq_units_both_ways).
% 0.14/0.35     dependent: set(dynamic_demod_all).
% 0.14/0.35     dependent: set(dynamic_demod).
% 0.14/0.35     dependent: set(order_eq).
% 0.14/0.35     dependent: set(back_demod).
% 0.14/0.35     dependent: set(lrpo).
% 0.14/0.35     dependent: set(hyper_res).
% 0.14/0.35     dependent: set(unit_deletion).
% 0.14/0.35     dependent: set(factor).
% 0.14/0.35  
% 0.14/0.35  ------------> process usable:
% 0.14/0.35  
% 0.14/0.35  ------------> process sos:
% 0.14/0.35    Following clause subsumed by 19 during input processing: 0 [copy,19,flip.1] {-} A=A.
% 0.14/0.35  19 back subsumes 14.
% 0.14/0.35  
% 0.14/0.35  ======= end of input processing =======
% 0.20/0.39  
% 0.20/0.39  Model 1 (0.00 seconds, 0 Inserts)
% 0.20/0.39  
% 0.20/0.39  Stopped by limit on number of solutions
% 0.20/0.39  
% 0.20/0.39  
% 0.20/0.39  -------------- Softie stats --------------
% 0.20/0.39  
% 0.20/0.39  UPDATE_STOP: 300
% 0.20/0.39  SFINDER_TIME_LIMIT: 2
% 0.20/0.39  SHORT_CLAUSE_CUTOFF: 4
% 0.20/0.39  number of clauses in intial UL: 13
% 0.20/0.39  number of clauses initially in problem: 18
% 0.20/0.39  percentage of clauses intially in UL: 72
% 0.20/0.39  percentage of distinct symbols occuring in initial UL: 88
% 0.20/0.39  percent of all initial clauses that are short: 100
% 0.20/0.39  absolute distinct symbol count: 9
% 0.20/0.39     distinct predicate count: 4
% 0.20/0.39     distinct function count: 3
% 0.20/0.39     distinct constant count: 2
% 0.20/0.39  
% 0.20/0.39  ---------- no more Softie stats ----------
% 0.20/0.39  
% 0.20/0.39  
% 0.20/0.39  
% 0.20/0.39  =========== start of search ===========
% 0.20/0.39  
% 0.20/0.39  -------- PROOF -------- 
% 0.20/0.39  % SZS status Theorem
% 0.20/0.39  % SZS output start Refutation
% 0.20/0.39  
% 0.20/0.39  ----> UNIT CONFLICT at   0.01 sec ----> 22 [binary,20.1,13.1] {-} $F.
% 0.20/0.39  
% 0.20/0.39  Length of proof is 1.  Level of proof is 1.
% 0.20/0.39  
% 0.20/0.39  ---------------- PROOF ----------------
% 0.20/0.39  % SZS status Theorem
% 0.20/0.39  % SZS output start Refutation
% 0.20/0.39  
% 0.20/0.39  2 [] {+} difference(A,B)=empty_set| -subset(A,B).
% 0.20/0.39  13 [] {+} difference(empty_set,$c1)!=empty_set.
% 0.20/0.39  15 [] {-} subset(empty_set,A).
% 0.20/0.39  20 [hyper,15,2] {-} difference(empty_set,A)=empty_set.
% 0.20/0.39  22 [binary,20.1,13.1] {-} $F.
% 0.20/0.39  
% 0.20/0.39  % SZS output end Refutation
% 0.20/0.39  ------------ end of proof -------------
% 0.20/0.39  
% 0.20/0.39  
% 0.20/0.39  Search stopped by max_proofs option.
% 0.20/0.39  
% 0.20/0.39  
% 0.20/0.39  Search stopped by max_proofs option.
% 0.20/0.39  
% 0.20/0.39  ============ end of search ============
% 0.20/0.39  
% 0.20/0.39  ----------- soft-scott stats ----------
% 0.20/0.39  
% 0.20/0.39  true clauses given           0      (0.0%)
% 0.20/0.39  false clauses given          1
% 0.20/0.39  
% 0.20/0.39        FALSE     TRUE
% 0.20/0.39     3  0         2
% 0.20/0.39     6  1         0
% 0.20/0.39     8  1         0
% 0.20/0.39  tot:  2         2      (50.0% true)
% 0.20/0.39  
% 0.20/0.39  
% 0.20/0.39  Model 1 (0.00 seconds, 0 Inserts)
% 0.20/0.39  
% 0.20/0.39  That finishes the proof of the theorem.
% 0.20/0.39  
% 0.20/0.39  Process 25898 finished Sun Jul 10 14:08:44 2022
%------------------------------------------------------------------------------