TSTP Solution File: SET872+1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : SET872+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n027.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 13:14:26 EDT 2022

% Result   : Theorem 1.69s 1.96s
% Output   : Refutation 1.69s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    4
%            Number of leaves      :    7
% Syntax   : Number of clauses     :   12 (   8 unt;   1 nHn;   8 RR)
%            Number of literals    :   18 (   7 equ;   7 neg)
%            Maximal clause size   :    3 (   1 avg)
%            Maximal term depth    :    3 (   1 avg)
%            Number of predicates  :    4 (   2 usr;   1 prp; 0-2 aty)
%            Number of functors    :    5 (   5 usr;   2 con; 0-2 aty)
%            Number of variables   :   16 (   2 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(2,axiom,
    ( A != singleton(B)
    | ~ in(C,A)
    | C = B ),
    file('SET872+1.p',unknown),
    [] ).

cnf(7,axiom,
    ( A != unordered_pair(B,C)
    | in(D,A)
    | D != C ),
    file('SET872+1.p',unknown),
    [] ).

cnf(11,axiom,
    ( subset(A,B)
    | ~ in(dollar_f3(A,B),B) ),
    file('SET872+1.p',unknown),
    [] ).

cnf(13,axiom,
    ~ subset(singleton(dollar_c4),unordered_pair(dollar_c4,dollar_c3)),
    file('SET872+1.p',unknown),
    [] ).

cnf(16,axiom,
    A = A,
    file('SET872+1.p',unknown),
    [] ).

cnf(17,axiom,
    unordered_pair(A,B) = unordered_pair(B,A),
    file('SET872+1.p',unknown),
    [] ).

cnf(20,axiom,
    ( subset(A,B)
    | in(dollar_f3(A,B),A) ),
    file('SET872+1.p',unknown),
    [] ).

cnf(24,plain,
    in(A,unordered_pair(B,A)),
    inference(hyper,[status(thm)],[16,7,16]),
    [iquote('hyper,16,7,16')] ).

cnf(58,plain,
    ~ subset(singleton(dollar_c4),unordered_pair(dollar_c3,dollar_c4)),
    inference(para_from,[status(thm),theory(equality)],[17,13]),
    [iquote('para_from,17.1.1,13.1.2')] ).

cnf(101,plain,
    in(dollar_f3(singleton(dollar_c4),unordered_pair(dollar_c3,dollar_c4)),singleton(dollar_c4)),
    inference(hyper,[status(thm)],[20,58]),
    [iquote('hyper,20,58')] ).

cnf(250,plain,
    dollar_f3(singleton(dollar_c4),unordered_pair(dollar_c3,dollar_c4)) = dollar_c4,
    inference(hyper,[status(thm)],[101,2,16]),
    [iquote('hyper,101,2,16')] ).

cnf(457,plain,
    $false,
    inference(unit_del,[status(thm)],[inference(para_from,[status(thm),theory(equality)],[250,11]),58,24]),
    [iquote('para_from,250.1.1,11.2.1,unit_del,58,24')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.09/0.11  % Problem  : SET872+1 : TPTP v8.1.0. Released v3.2.0.
% 0.09/0.12  % Command  : otter-tptp-script %s
% 0.12/0.33  % Computer : n027.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 10:55:36 EDT 2022
% 0.12/0.33  % CPUTime  : 
% 1.69/1.93  ----- Otter 3.3f, August 2004 -----
% 1.69/1.93  The process was started by sandbox2 on n027.cluster.edu,
% 1.69/1.93  Wed Jul 27 10:55:37 2022
% 1.69/1.93  The command was "./otter".  The process ID is 2006.
% 1.69/1.93  
% 1.69/1.93  set(prolog_style_variables).
% 1.69/1.93  set(auto).
% 1.69/1.93     dependent: set(auto1).
% 1.69/1.93     dependent: set(process_input).
% 1.69/1.93     dependent: clear(print_kept).
% 1.69/1.93     dependent: clear(print_new_demod).
% 1.69/1.93     dependent: clear(print_back_demod).
% 1.69/1.93     dependent: clear(print_back_sub).
% 1.69/1.93     dependent: set(control_memory).
% 1.69/1.93     dependent: assign(max_mem, 12000).
% 1.69/1.93     dependent: assign(pick_given_ratio, 4).
% 1.69/1.93     dependent: assign(stats_level, 1).
% 1.69/1.93     dependent: assign(max_seconds, 10800).
% 1.69/1.93  clear(print_given).
% 1.69/1.93  
% 1.69/1.93  formula_list(usable).
% 1.69/1.93  all A (A=A).
% 1.69/1.93  all A B (in(A,B)-> -in(B,A)).
% 1.69/1.93  all A B (unordered_pair(A,B)=unordered_pair(B,A)).
% 1.69/1.93  all A B (B=singleton(A)<-> (all C (in(C,B)<->C=A))).
% 1.69/1.93  all A B C (C=unordered_pair(A,B)<-> (all D (in(D,C)<->D=A|D=B))).
% 1.69/1.93  all A B (subset(A,B)<-> (all C (in(C,A)->in(C,B)))).
% 1.69/1.93  exists A empty(A).
% 1.69/1.93  exists A (-empty(A)).
% 1.69/1.93  all A B subset(A,A).
% 1.69/1.93  -(all A B subset(singleton(A),unordered_pair(A,B))).
% 1.69/1.93  end_of_list.
% 1.69/1.93  
% 1.69/1.93  -------> usable clausifies to:
% 1.69/1.93  
% 1.69/1.93  list(usable).
% 1.69/1.93  0 [] A=A.
% 1.69/1.93  0 [] -in(A,B)| -in(B,A).
% 1.69/1.93  0 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.69/1.93  0 [] B!=singleton(A)| -in(C,B)|C=A.
% 1.69/1.93  0 [] B!=singleton(A)|in(C,B)|C!=A.
% 1.69/1.93  0 [] B=singleton(A)|in($f1(A,B),B)|$f1(A,B)=A.
% 1.69/1.93  0 [] B=singleton(A)| -in($f1(A,B),B)|$f1(A,B)!=A.
% 1.69/1.93  0 [] C!=unordered_pair(A,B)| -in(D,C)|D=A|D=B.
% 1.69/1.93  0 [] C!=unordered_pair(A,B)|in(D,C)|D!=A.
% 1.69/1.93  0 [] C!=unordered_pair(A,B)|in(D,C)|D!=B.
% 1.69/1.93  0 [] C=unordered_pair(A,B)|in($f2(A,B,C),C)|$f2(A,B,C)=A|$f2(A,B,C)=B.
% 1.69/1.93  0 [] C=unordered_pair(A,B)| -in($f2(A,B,C),C)|$f2(A,B,C)!=A.
% 1.69/1.93  0 [] C=unordered_pair(A,B)| -in($f2(A,B,C),C)|$f2(A,B,C)!=B.
% 1.69/1.93  0 [] -subset(A,B)| -in(C,A)|in(C,B).
% 1.69/1.93  0 [] subset(A,B)|in($f3(A,B),A).
% 1.69/1.93  0 [] subset(A,B)| -in($f3(A,B),B).
% 1.69/1.93  0 [] empty($c1).
% 1.69/1.93  0 [] -empty($c2).
% 1.69/1.93  0 [] subset(A,A).
% 1.69/1.93  0 [] -subset(singleton($c4),unordered_pair($c4,$c3)).
% 1.69/1.93  end_of_list.
% 1.69/1.93  
% 1.69/1.93  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=4.
% 1.69/1.93  
% 1.69/1.93  This ia a non-Horn set with equality.  The strategy will be
% 1.69/1.93  Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.69/1.93  deletion, with positive clauses in sos and nonpositive
% 1.69/1.93  clauses in usable.
% 1.69/1.93  
% 1.69/1.93     dependent: set(knuth_bendix).
% 1.69/1.93     dependent: set(anl_eq).
% 1.69/1.93     dependent: set(para_from).
% 1.69/1.93     dependent: set(para_into).
% 1.69/1.93     dependent: clear(para_from_right).
% 1.69/1.93     dependent: clear(para_into_right).
% 1.69/1.93     dependent: set(para_from_vars).
% 1.69/1.93     dependent: set(eq_units_both_ways).
% 1.69/1.93     dependent: set(dynamic_demod_all).
% 1.69/1.93     dependent: set(dynamic_demod).
% 1.69/1.93     dependent: set(order_eq).
% 1.69/1.93     dependent: set(back_demod).
% 1.69/1.93     dependent: set(lrpo).
% 1.69/1.93     dependent: set(hyper_res).
% 1.69/1.93     dependent: set(unit_deletion).
% 1.69/1.93     dependent: set(factor).
% 1.69/1.93  
% 1.69/1.93  ------------> process usable:
% 1.69/1.93  ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.69/1.93  ** KEPT (pick-wt=10): 2 [] A!=singleton(B)| -in(C,A)|C=B.
% 1.69/1.93  ** KEPT (pick-wt=10): 3 [] A!=singleton(B)|in(C,A)|C!=B.
% 1.69/1.93  ** KEPT (pick-wt=14): 4 [] A=singleton(B)| -in($f1(B,A),A)|$f1(B,A)!=B.
% 1.69/1.93  ** KEPT (pick-wt=14): 5 [] A!=unordered_pair(B,C)| -in(D,A)|D=B|D=C.
% 1.69/1.93  ** KEPT (pick-wt=11): 6 [] A!=unordered_pair(B,C)|in(D,A)|D!=B.
% 1.69/1.93  ** KEPT (pick-wt=11): 7 [] A!=unordered_pair(B,C)|in(D,A)|D!=C.
% 1.69/1.93  ** KEPT (pick-wt=17): 8 [] A=unordered_pair(B,C)| -in($f2(B,C,A),A)|$f2(B,C,A)!=B.
% 1.69/1.93  ** KEPT (pick-wt=17): 9 [] A=unordered_pair(B,C)| -in($f2(B,C,A),A)|$f2(B,C,A)!=C.
% 1.69/1.93  ** KEPT (pick-wt=9): 10 [] -subset(A,B)| -in(C,A)|in(C,B).
% 1.69/1.93  ** KEPT (pick-wt=8): 11 [] subset(A,B)| -in($f3(A,B),B).
% 1.69/1.93  ** KEPT (pick-wt=2): 12 [] -empty($c2).
% 1.69/1.93  ** KEPT (pick-wt=6): 13 [] -subset(singleton($c4),unordered_pair($c4,$c3)).
% 1.69/1.93  
% 1.69/1.93  ------------> process sos:
% 1.69/1.93  ** KEPT (pick-wt=3): 16 [] A=A.
% 1.69/1.93  ** KEPT (pick-wt=7): 17 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.69/1.93  ** KEPT (pick-wt=14): 18 [] A=singleton(B)|in($f1(B,A),A)|$f1(B,A)=B.
% 1.69/1.93  ** KEPT (pick-wt=23): 19 [] A=unordered_pair(B,C)|in($f2(B,C,A),A)|$f2(B,C,A)=B|$f2(B,C,A)=C.
% 1.69/1.93  ** KEPT (pick-wt=8): 20 [] subset(A,B)|in($f3(A,B),A).
% 1.69/1.93  ** KEPT (pick-wt=2): 21 [] empty($c1).
% 1.69/1.93  ** KEPT (pick-wt=3): 22 [] subset(A,A).
% 1.69/1.93    Following clause subsumed by 16 during input processing: 0 [copy,16,flip.1] A=A.
% 1.69/1.93    Following clause subsumed by 17 during input processing: 0 [copy,17,flip.1] unordered_pair(A,B)=unordered_pair(B,A).
% 1.69/1.96  
% 1.69/1.96  ======= end of input processing =======
% 1.69/1.96  
% 1.69/1.96  =========== start of search ===========
% 1.69/1.96  
% 1.69/1.96  -------- PROOF -------- 
% 1.69/1.96  
% 1.69/1.96  -----> EMPTY CLAUSE at   0.04 sec ----> 457 [para_from,250.1.1,11.2.1,unit_del,58,24] $F.
% 1.69/1.96  
% 1.69/1.96  Length of proof is 4.  Level of proof is 3.
% 1.69/1.96  
% 1.69/1.96  ---------------- PROOF ----------------
% 1.69/1.96  % SZS status Theorem
% 1.69/1.96  % SZS output start Refutation
% See solution above
% 1.69/1.96  ------------ end of proof -------------
% 1.69/1.96  
% 1.69/1.96  
% 1.69/1.96  Search stopped by max_proofs option.
% 1.69/1.96  
% 1.69/1.96  
% 1.69/1.96  Search stopped by max_proofs option.
% 1.69/1.96  
% 1.69/1.96  ============ end of search ============
% 1.69/1.96  
% 1.69/1.96  -------------- statistics -------------
% 1.69/1.96  clauses given                 17
% 1.69/1.96  clauses generated            582
% 1.69/1.96  clauses kept                 454
% 1.69/1.96  clauses forward subsumed     145
% 1.69/1.96  clauses back subsumed          0
% 1.69/1.96  Kbytes malloced             1953
% 1.69/1.96  
% 1.69/1.96  ----------- times (seconds) -----------
% 1.69/1.96  user CPU time          0.04          (0 hr, 0 min, 0 sec)
% 1.69/1.96  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.69/1.96  wall-clock time        1             (0 hr, 0 min, 1 sec)
% 1.69/1.96  
% 1.69/1.96  That finishes the proof of the theorem.
% 1.69/1.96  
% 1.69/1.96  Process 2006 finished Wed Jul 27 10:55:38 2022
% 1.69/1.96  Otter interrupted
% 1.69/1.96  PROOF FOUND
%------------------------------------------------------------------------------