TSTP Solution File: SET900+1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : SET900+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n015.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 13:14:29 EDT 2022

% Result   : Theorem 1.57s 2.14s
% Output   : Refutation 1.57s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    3
%            Number of leaves      :    5
% Syntax   : Number of clauses     :    9 (   5 unt;   3 nHn;   7 RR)
%            Number of literals    :   15 (  12 equ;   6 neg)
%            Maximal clause size   :    3 (   1 avg)
%            Maximal term depth    :    2 (   1 avg)
%            Number of predicates  :    3 (   1 usr;   1 prp; 0-2 aty)
%            Number of functors    :    5 (   5 usr;   3 con; 0-2 aty)
%            Number of variables   :    6 (   0 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(3,axiom,
    dollar_c4 != singleton(dollar_c3),
    file('SET900+1.p',unknown),
    [] ).

cnf(4,plain,
    singleton(dollar_c3) != dollar_c4,
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[3])]),
    [iquote('copy,3,flip.1')] ).

cnf(5,axiom,
    dollar_c4 != empty_set,
    file('SET900+1.p',unknown),
    [] ).

cnf(6,plain,
    empty_set != dollar_c4,
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[5])]),
    [iquote('copy,5,flip.1')] ).

cnf(7,axiom,
    ( ~ in(A,dollar_c4)
    | A = dollar_c3 ),
    file('SET900+1.p',unknown),
    [] ).

cnf(8,axiom,
    ( A = singleton(B)
    | A = empty_set
    | dollar_f1(A,B) != B ),
    file('SET900+1.p',unknown),
    [] ).

cnf(13,axiom,
    ( A = singleton(B)
    | A = empty_set
    | in(dollar_f1(A,B),A) ),
    file('SET900+1.p',unknown),
    [] ).

cnf(22,plain,
    ( singleton(A) = dollar_c4
    | dollar_f1(dollar_c4,A) = dollar_c3 ),
    inference(flip,[status(thm),theory(equality)],[inference(unit_del,[status(thm)],[inference(hyper,[status(thm)],[13,7]),6])]),
    [iquote('hyper,13,7,unit_del,6,flip.1')] ).

cnf(45,plain,
    $false,
    inference(unit_del,[status(thm)],[inference(hyper,[status(thm)],[22,8]),4,4,6]),
    [iquote('hyper,22,8,unit_del,4,4,6')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12  % Problem  : SET900+1 : TPTP v8.1.0. Released v3.2.0.
% 0.06/0.12  % Command  : otter-tptp-script %s
% 0.12/0.33  % Computer : n015.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 : Wed Jul 27 10:41:42 EDT 2022
% 0.12/0.34  % CPUTime  : 
% 1.57/2.14  ----- Otter 3.3f, August 2004 -----
% 1.57/2.14  The process was started by sandbox2 on n015.cluster.edu,
% 1.57/2.14  Wed Jul 27 10:41:42 2022
% 1.57/2.14  The command was "./otter".  The process ID is 2920.
% 1.57/2.14  
% 1.57/2.14  set(prolog_style_variables).
% 1.57/2.14  set(auto).
% 1.57/2.14     dependent: set(auto1).
% 1.57/2.14     dependent: set(process_input).
% 1.57/2.14     dependent: clear(print_kept).
% 1.57/2.14     dependent: clear(print_new_demod).
% 1.57/2.14     dependent: clear(print_back_demod).
% 1.57/2.14     dependent: clear(print_back_sub).
% 1.57/2.14     dependent: set(control_memory).
% 1.57/2.14     dependent: assign(max_mem, 12000).
% 1.57/2.14     dependent: assign(pick_given_ratio, 4).
% 1.57/2.14     dependent: assign(stats_level, 1).
% 1.57/2.14     dependent: assign(max_seconds, 10800).
% 1.57/2.14  clear(print_given).
% 1.57/2.14  
% 1.57/2.14  formula_list(usable).
% 1.57/2.14  all A (A=A).
% 1.57/2.14  all A B (in(A,B)-> -in(B,A)).
% 1.57/2.14  empty(empty_set).
% 1.57/2.14  exists A empty(A).
% 1.57/2.14  exists A (-empty(A)).
% 1.57/2.14  -(all A B (-(A!=singleton(B)&A!=empty_set& (all C (-(in(C,A)&C!=B)))))).
% 1.57/2.14  all A B (-(A!=singleton(B)&A!=empty_set& (all C (-(in(C,A)&C!=B))))).
% 1.57/2.14  end_of_list.
% 1.57/2.14  
% 1.57/2.14  -------> usable clausifies to:
% 1.57/2.14  
% 1.57/2.14  list(usable).
% 1.57/2.14  0 [] A=A.
% 1.57/2.14  0 [] -in(A,B)| -in(B,A).
% 1.57/2.14  0 [] empty(empty_set).
% 1.57/2.14  0 [] empty($c1).
% 1.57/2.14  0 [] -empty($c2).
% 1.57/2.14  0 [] $c4!=singleton($c3).
% 1.57/2.14  0 [] $c4!=empty_set.
% 1.57/2.14  0 [] -in(C,$c4)|C=$c3.
% 1.57/2.14  0 [] A=singleton(B)|A=empty_set|in($f1(A,B),A).
% 1.57/2.14  0 [] A=singleton(B)|A=empty_set|$f1(A,B)!=B.
% 1.57/2.14  end_of_list.
% 1.57/2.14  
% 1.57/2.14  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.57/2.14  
% 1.57/2.14  This ia a non-Horn set with equality.  The strategy will be
% 1.57/2.14  Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.57/2.14  deletion, with positive clauses in sos and nonpositive
% 1.57/2.14  clauses in usable.
% 1.57/2.14  
% 1.57/2.14     dependent: set(knuth_bendix).
% 1.57/2.14     dependent: set(anl_eq).
% 1.57/2.14     dependent: set(para_from).
% 1.57/2.14     dependent: set(para_into).
% 1.57/2.14     dependent: clear(para_from_right).
% 1.57/2.14     dependent: clear(para_into_right).
% 1.57/2.14     dependent: set(para_from_vars).
% 1.57/2.14     dependent: set(eq_units_both_ways).
% 1.57/2.14     dependent: set(dynamic_demod_all).
% 1.57/2.14     dependent: set(dynamic_demod).
% 1.57/2.14     dependent: set(order_eq).
% 1.57/2.14     dependent: set(back_demod).
% 1.57/2.14     dependent: set(lrpo).
% 1.57/2.14     dependent: set(hyper_res).
% 1.57/2.14     dependent: set(unit_deletion).
% 1.57/2.14     dependent: set(factor).
% 1.57/2.14  
% 1.57/2.14  ------------> process usable:
% 1.57/2.14  ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.57/2.14  ** KEPT (pick-wt=2): 2 [] -empty($c2).
% 1.57/2.14  ** KEPT (pick-wt=4): 4 [copy,3,flip.1] singleton($c3)!=$c4.
% 1.57/2.14  ** KEPT (pick-wt=3): 6 [copy,5,flip.1] empty_set!=$c4.
% 1.57/2.14  ** KEPT (pick-wt=6): 7 [] -in(A,$c4)|A=$c3.
% 1.57/2.14  ** KEPT (pick-wt=12): 8 [] A=singleton(B)|A=empty_set|$f1(A,B)!=B.
% 1.57/2.14  
% 1.57/2.14  ------------> process sos:
% 1.57/2.14  ** KEPT (pick-wt=3): 10 [] A=A.
% 1.57/2.14  ** KEPT (pick-wt=2): 11 [] empty(empty_set).
% 1.57/2.14  ** KEPT (pick-wt=2): 12 [] empty($c1).
% 1.57/2.14  ** KEPT (pick-wt=12): 13 [] A=singleton(B)|A=empty_set|in($f1(A,B),A).
% 1.57/2.14    Following clause subsumed by 10 during input processing: 0 [copy,10,flip.1] A=A.
% 1.57/2.14  
% 1.57/2.14  ======= end of input processing =======
% 1.57/2.14  
% 1.57/2.14  =========== start of search ===========
% 1.57/2.14  
% 1.57/2.14  -------- PROOF -------- 
% 1.57/2.14  
% 1.57/2.14  -----> EMPTY CLAUSE at   0.00 sec ----> 45 [hyper,22,8,unit_del,4,4,6] $F.
% 1.57/2.14  
% 1.57/2.14  Length of proof is 3.  Level of proof is 2.
% 1.57/2.14  
% 1.57/2.14  ---------------- PROOF ----------------
% 1.57/2.14  % SZS status Theorem
% 1.57/2.14  % SZS output start Refutation
% See solution above
% 1.57/2.14  ------------ end of proof -------------
% 1.57/2.14  
% 1.57/2.14  
% 1.57/2.14  Search stopped by max_proofs option.
% 1.57/2.14  
% 1.57/2.14  
% 1.57/2.14  Search stopped by max_proofs option.
% 1.57/2.14  
% 1.57/2.14  ============ end of search ============
% 1.57/2.14  
% 1.57/2.14  -------------- statistics -------------
% 1.57/2.14  clauses given                  7
% 1.57/2.14  clauses generated             63
% 1.57/2.14  clauses kept                  41
% 1.57/2.14  clauses forward subsumed      32
% 1.57/2.14  clauses back subsumed          0
% 1.57/2.14  Kbytes malloced              976
% 1.57/2.14  
% 1.57/2.14  ----------- times (seconds) -----------
% 1.57/2.14  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.57/2.14  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.57/2.14  wall-clock time        1             (0 hr, 0 min, 1 sec)
% 1.57/2.14  
% 1.57/2.14  That finishes the proof of the theorem.
% 1.57/2.14  
% 1.57/2.14  Process 2920 finished Wed Jul 27 10:41:43 2022
% 1.57/2.14  Otter interrupted
% 1.57/2.14  PROOF FOUND
%------------------------------------------------------------------------------