TSTP Solution File: SET063+3 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : SET063+3 : TPTP v8.1.0. Released v2.2.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n016.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 : Wed Jul 27 13:12:58 EDT 2022

% Result   : Theorem 1.71s 1.93s
% Output   : Refutation 1.71s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    2
%            Number of leaves      :    4
% Syntax   : Number of clauses     :    7 (   6 unt;   0 nHn;   6 RR)
%            Number of literals    :    9 (   4 equ;   4 neg)
%            Maximal clause size   :    3 (   1 avg)
%            Maximal term depth    :    1 (   1 avg)
%            Number of predicates  :    3 (   1 usr;   1 prp; 0-2 aty)
%            Number of functors    :    2 (   2 usr;   2 con; 0-0 aty)
%            Number of variables   :    3 (   1 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(6,axiom,
    ( A = B
    | ~ subset(A,B)
    | ~ subset(B,A) ),
    file('SET063+3.p',unknown),
    [] ).

cnf(8,axiom,
    dollar_c1 != empty_set,
    file('SET063+3.p',unknown),
    [] ).

cnf(9,plain,
    empty_set != dollar_c1,
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[8])]),
    [iquote('copy,8,flip.1')] ).

cnf(12,axiom,
    subset(empty_set,A),
    file('SET063+3.p',unknown),
    [] ).

cnf(16,axiom,
    subset(dollar_c1,empty_set),
    file('SET063+3.p',unknown),
    [] ).

cnf(18,plain,
    empty_set = dollar_c1,
    inference(hyper,[status(thm)],[16,6,12]),
    [iquote('hyper,16,6,12')] ).

cnf(20,plain,
    $false,
    inference(binary,[status(thm)],[18,9]),
    [iquote('binary,18.1,9.1')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.11  % Problem  : SET063+3 : TPTP v8.1.0. Released v2.2.0.
% 0.07/0.12  % Command  : otter-tptp-script %s
% 0.13/0.33  % Computer : n016.cluster.edu
% 0.13/0.33  % Model    : x86_64 x86_64
% 0.13/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33  % Memory   : 8042.1875MB
% 0.13/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33  % CPULimit : 300
% 0.13/0.33  % WCLimit  : 300
% 0.13/0.33  % DateTime : Wed Jul 27 11:07:22 EDT 2022
% 0.13/0.33  % CPUTime  : 
% 1.71/1.93  ----- Otter 3.3f, August 2004 -----
% 1.71/1.93  The process was started by sandbox on n016.cluster.edu,
% 1.71/1.93  Wed Jul 27 11:07:22 2022
% 1.71/1.93  The command was "./otter".  The process ID is 24939.
% 1.71/1.93  
% 1.71/1.93  set(prolog_style_variables).
% 1.71/1.93  set(auto).
% 1.71/1.93     dependent: set(auto1).
% 1.71/1.93     dependent: set(process_input).
% 1.71/1.93     dependent: clear(print_kept).
% 1.71/1.93     dependent: clear(print_new_demod).
% 1.71/1.93     dependent: clear(print_back_demod).
% 1.71/1.93     dependent: clear(print_back_sub).
% 1.71/1.93     dependent: set(control_memory).
% 1.71/1.93     dependent: assign(max_mem, 12000).
% 1.71/1.93     dependent: assign(pick_given_ratio, 4).
% 1.71/1.93     dependent: assign(stats_level, 1).
% 1.71/1.93     dependent: assign(max_seconds, 10800).
% 1.71/1.93  clear(print_given).
% 1.71/1.93  
% 1.71/1.93  formula_list(usable).
% 1.71/1.93  all A (A=A).
% 1.71/1.93  all B subset(empty_set,B).
% 1.71/1.93  all B (-member(B,empty_set)).
% 1.71/1.93  all B C (subset(B,C)<-> (all D (member(D,B)->member(D,C)))).
% 1.71/1.93  all B C (B=C<->subset(B,C)&subset(C,B)).
% 1.71/1.93  all B subset(B,B).
% 1.71/1.93  all B (empty(B)<-> (all C (-member(C,B)))).
% 1.71/1.93  -(all B (subset(B,empty_set)->B=empty_set)).
% 1.71/1.93  end_of_list.
% 1.71/1.93  
% 1.71/1.93  -------> usable clausifies to:
% 1.71/1.93  
% 1.71/1.93  list(usable).
% 1.71/1.93  0 [] A=A.
% 1.71/1.93  0 [] subset(empty_set,B).
% 1.71/1.93  0 [] -member(B,empty_set).
% 1.71/1.93  0 [] -subset(B,C)| -member(D,B)|member(D,C).
% 1.71/1.93  0 [] subset(B,C)|member($f1(B,C),B).
% 1.71/1.93  0 [] subset(B,C)| -member($f1(B,C),C).
% 1.71/1.93  0 [] B!=C|subset(B,C).
% 1.71/1.93  0 [] B!=C|subset(C,B).
% 1.71/1.93  0 [] B=C| -subset(B,C)| -subset(C,B).
% 1.71/1.93  0 [] subset(B,B).
% 1.71/1.93  0 [] -empty(B)| -member(C,B).
% 1.71/1.93  0 [] empty(B)|member($f2(B),B).
% 1.71/1.93  0 [] subset($c1,empty_set).
% 1.71/1.93  0 [] $c1!=empty_set.
% 1.71/1.93  end_of_list.
% 1.71/1.93  
% 1.71/1.93  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.71/1.93  
% 1.71/1.93  This ia a non-Horn set with equality.  The strategy will be
% 1.71/1.93  Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.71/1.93  deletion, with positive clauses in sos and nonpositive
% 1.71/1.93  clauses in usable.
% 1.71/1.93  
% 1.71/1.93     dependent: set(knuth_bendix).
% 1.71/1.93     dependent: set(anl_eq).
% 1.71/1.93     dependent: set(para_from).
% 1.71/1.93     dependent: set(para_into).
% 1.71/1.93     dependent: clear(para_from_right).
% 1.71/1.93     dependent: clear(para_into_right).
% 1.71/1.93     dependent: set(para_from_vars).
% 1.71/1.93     dependent: set(eq_units_both_ways).
% 1.71/1.93     dependent: set(dynamic_demod_all).
% 1.71/1.93     dependent: set(dynamic_demod).
% 1.71/1.93     dependent: set(order_eq).
% 1.71/1.93     dependent: set(back_demod).
% 1.71/1.93     dependent: set(lrpo).
% 1.71/1.93     dependent: set(hyper_res).
% 1.71/1.93     dependent: set(unit_deletion).
% 1.71/1.93     dependent: set(factor).
% 1.71/1.93  
% 1.71/1.93  ------------> process usable:
% 1.71/1.93  ** KEPT (pick-wt=3): 1 [] -member(A,empty_set).
% 1.71/1.93  ** KEPT (pick-wt=9): 2 [] -subset(A,B)| -member(C,A)|member(C,B).
% 1.71/1.93  ** KEPT (pick-wt=8): 3 [] subset(A,B)| -member($f1(A,B),B).
% 1.71/1.93  ** KEPT (pick-wt=6): 4 [] A!=B|subset(A,B).
% 1.71/1.93  ** KEPT (pick-wt=6): 5 [] A!=B|subset(B,A).
% 1.71/1.93  ** KEPT (pick-wt=9): 6 [] A=B| -subset(A,B)| -subset(B,A).
% 1.71/1.93  ** KEPT (pick-wt=5): 7 [] -empty(A)| -member(B,A).
% 1.71/1.93  ** KEPT (pick-wt=3): 9 [copy,8,flip.1] empty_set!=$c1.
% 1.71/1.93  
% 1.71/1.93  ------------> process sos:
% 1.71/1.93  ** KEPT (pick-wt=3): 11 [] A=A.
% 1.71/1.93  ** KEPT (pick-wt=3): 12 [] subset(empty_set,A).
% 1.71/1.93  ** KEPT (pick-wt=8): 13 [] subset(A,B)|member($f1(A,B),A).
% 1.71/1.93  ** KEPT (pick-wt=3): 14 [] subset(A,A).
% 1.71/1.93  ** KEPT (pick-wt=6): 15 [] empty(A)|member($f2(A),A).
% 1.71/1.93  ** KEPT (pick-wt=3): 16 [] subset($c1,empty_set).
% 1.71/1.93    Following clause subsumed by 11 during input processing: 0 [copy,11,flip.1] A=A.
% 1.71/1.93  11 back subsumes 10.
% 1.71/1.93  
% 1.71/1.93  ======= end of input processing =======
% 1.71/1.93  
% 1.71/1.93  =========== start of search ===========
% 1.71/1.93  
% 1.71/1.93  -------- PROOF -------- 
% 1.71/1.93  
% 1.71/1.93  ----> UNIT CONFLICT at   0.00 sec ----> 20 [binary,18.1,9.1] $F.
% 1.71/1.93  
% 1.71/1.93  Length of proof is 2.  Level of proof is 1.
% 1.71/1.93  
% 1.71/1.93  ---------------- PROOF ----------------
% 1.71/1.93  % SZS status Theorem
% 1.71/1.93  % SZS output start Refutation
% See solution above
% 1.71/1.93  ------------ end of proof -------------
% 1.71/1.93  
% 1.71/1.93  
% 1.71/1.93  Search stopped by max_proofs option.
% 1.71/1.93  
% 1.71/1.93  
% 1.71/1.93  Search stopped by max_proofs option.
% 1.71/1.93  
% 1.71/1.93  ============ end of search ============
% 1.71/1.93  
% 1.71/1.93  -------------- statistics -------------
% 1.71/1.93  clauses given                  4
% 1.71/1.93  clauses generated             11
% 1.71/1.93  clauses kept                  17
% 1.71/1.93  clauses forward subsumed       9
% 1.71/1.93  clauses back subsumed          1
% 1.71/1.93  Kbytes malloced              976
% 1.71/1.93  
% 1.71/1.93  ----------- times (seconds) -----------
% 1.71/1.93  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.71/1.93  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.71/1.93  wall-clock time        2             (0 hr, 0 min, 2 sec)
% 1.71/1.93  
% 1.71/1.93  That finishes the proof of the theorem.
% 1.71/1.93  
% 1.71/1.93  Process 24939 finished Wed Jul 27 11:07:24 2022
% 1.71/1.93  Otter interrupted
% 1.71/1.93  PROOF FOUND
%------------------------------------------------------------------------------