TSTP Solution File: SEU160+3 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : SEU160+3 : TPTP v8.1.0. Released v3.2.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n019.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 13:14:59 EDT 2022

% Result   : Theorem 1.68s 1.89s
% Output   : Refutation 1.68s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    6
%            Number of leaves      :    7
% Syntax   : Number of clauses     :   16 (   6 unt;   4 nHn;  11 RR)
%            Number of literals    :   29 (  16 equ;  11 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    :    4 (   4 usr;   3 con; 0-1 aty)
%            Number of variables   :    8 (   3 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(2,axiom,
    ( ~ subset(dollar_c4,singleton(dollar_c3))
    | dollar_c4 != empty_set ),
    file('SEU160+3.p',unknown),
    [] ).

cnf(3,plain,
    ( ~ subset(dollar_c4,singleton(dollar_c3))
    | empty_set != dollar_c4 ),
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[2])]),
    [iquote('copy,2,flip.2')] ).

cnf(4,axiom,
    ( ~ subset(dollar_c4,singleton(dollar_c3))
    | dollar_c4 != singleton(dollar_c3) ),
    file('SEU160+3.p',unknown),
    [] ).

cnf(5,plain,
    ( ~ subset(dollar_c4,singleton(dollar_c3))
    | singleton(dollar_c3) != dollar_c4 ),
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[4])]),
    [iquote('copy,4,flip.2')] ).

cnf(6,axiom,
    ( ~ subset(A,singleton(B))
    | A = empty_set
    | A = singleton(B) ),
    file('SEU160+3.p',unknown),
    [] ).

cnf(7,axiom,
    ( subset(A,singleton(B))
    | A != empty_set ),
    file('SEU160+3.p',unknown),
    [] ).

cnf(9,axiom,
    A = A,
    file('SEU160+3.p',unknown),
    [] ).

cnf(10,axiom,
    subset(A,A),
    file('SEU160+3.p',unknown),
    [] ).

cnf(13,axiom,
    ( subset(dollar_c4,singleton(dollar_c3))
    | dollar_c4 = empty_set
    | dollar_c4 = singleton(dollar_c3) ),
    file('SEU160+3.p',unknown),
    [] ).

cnf(14,plain,
    ( subset(dollar_c4,singleton(dollar_c3))
    | empty_set = dollar_c4
    | singleton(dollar_c3) = dollar_c4 ),
    inference(flip,[status(thm),theory(equality)],[inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[13])])]),
    [iquote('copy,13,flip.2,flip.3')] ).

cnf(15,plain,
    subset(empty_set,singleton(A)),
    inference(hyper,[status(thm)],[9,7]),
    [iquote('hyper,9,7')] ).

cnf(17,plain,
    ( empty_set = dollar_c4
    | singleton(dollar_c3) = dollar_c4 ),
    inference(factor_simp,[status(thm)],[inference(factor_simp,[status(thm)],[inference(hyper,[status(thm)],[14,6])])]),
    [iquote('hyper,14,6,factor_simp,factor_simp')] ).

cnf(29,plain,
    ( ~ subset(dollar_c4,singleton(dollar_c3))
    | empty_set = dollar_c4 ),
    inference(unit_del,[status(thm)],[inference(para_from,[status(thm),theory(equality)],[17,5]),9]),
    [iquote('para_from,17.2.1,5.2.1,unit_del,9')] ).

cnf(69,plain,
    empty_set = dollar_c4,
    inference(factor_simp,[status(thm)],[inference(unit_del,[status(thm)],[inference(para_into,[status(thm),theory(equality)],[29,17]),10])]),
    [iquote('para_into,29.1.2,17.2.1,unit_del,10,factor_simp')] ).

cnf(76,plain,
    subset(dollar_c4,singleton(A)),
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[15]),69]),
    [iquote('back_demod,15,demod,69')] ).

cnf(79,plain,
    $false,
    inference(unit_del,[status(thm)],[inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[3]),69]),76,9]),
    [iquote('back_demod,3,demod,69,unit_del,76,9')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.11  % Problem  : SEU160+3 : TPTP v8.1.0. Released v3.2.0.
% 0.07/0.12  % Command  : otter-tptp-script %s
% 0.12/0.33  % Computer : n019.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 07:42:37 EDT 2022
% 0.12/0.33  % CPUTime  : 
% 1.68/1.89  ----- Otter 3.3f, August 2004 -----
% 1.68/1.89  The process was started by sandbox on n019.cluster.edu,
% 1.68/1.89  Wed Jul 27 07:42:37 2022
% 1.68/1.89  The command was "./otter".  The process ID is 7487.
% 1.68/1.89  
% 1.68/1.89  set(prolog_style_variables).
% 1.68/1.89  set(auto).
% 1.68/1.89     dependent: set(auto1).
% 1.68/1.89     dependent: set(process_input).
% 1.68/1.89     dependent: clear(print_kept).
% 1.68/1.89     dependent: clear(print_new_demod).
% 1.68/1.89     dependent: clear(print_back_demod).
% 1.68/1.89     dependent: clear(print_back_sub).
% 1.68/1.89     dependent: set(control_memory).
% 1.68/1.89     dependent: assign(max_mem, 12000).
% 1.68/1.89     dependent: assign(pick_given_ratio, 4).
% 1.68/1.89     dependent: assign(stats_level, 1).
% 1.68/1.89     dependent: assign(max_seconds, 10800).
% 1.68/1.89  clear(print_given).
% 1.68/1.89  
% 1.68/1.89  formula_list(usable).
% 1.68/1.89  all A (A=A).
% 1.68/1.89  all A B subset(A,A).
% 1.68/1.89  empty(empty_set).
% 1.68/1.89  exists A empty(A).
% 1.68/1.89  exists A (-empty(A)).
% 1.68/1.89  -(all A B (subset(A,singleton(B))<->A=empty_set|A=singleton(B))).
% 1.68/1.89  all A B (subset(A,singleton(B))<->A=empty_set|A=singleton(B)).
% 1.68/1.89  end_of_list.
% 1.68/1.89  
% 1.68/1.89  -------> usable clausifies to:
% 1.68/1.89  
% 1.68/1.89  list(usable).
% 1.68/1.89  0 [] A=A.
% 1.68/1.89  0 [] subset(A,A).
% 1.68/1.89  0 [] empty(empty_set).
% 1.68/1.89  0 [] empty($c1).
% 1.68/1.89  0 [] -empty($c2).
% 1.68/1.89  0 [] subset($c4,singleton($c3))|$c4=empty_set|$c4=singleton($c3).
% 1.68/1.89  0 [] -subset($c4,singleton($c3))|$c4!=empty_set.
% 1.68/1.89  0 [] -subset($c4,singleton($c3))|$c4!=singleton($c3).
% 1.68/1.89  0 [] -subset(A,singleton(B))|A=empty_set|A=singleton(B).
% 1.68/1.89  0 [] subset(A,singleton(B))|A!=empty_set.
% 1.68/1.89  0 [] subset(A,singleton(B))|A!=singleton(B).
% 1.68/1.89  end_of_list.
% 1.68/1.89  
% 1.68/1.89  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.68/1.89  
% 1.68/1.89  This ia a non-Horn set with equality.  The strategy will be
% 1.68/1.89  Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.68/1.89  deletion, with positive clauses in sos and nonpositive
% 1.68/1.89  clauses in usable.
% 1.68/1.89  
% 1.68/1.89     dependent: set(knuth_bendix).
% 1.68/1.89     dependent: set(anl_eq).
% 1.68/1.89     dependent: set(para_from).
% 1.68/1.89     dependent: set(para_into).
% 1.68/1.89     dependent: clear(para_from_right).
% 1.68/1.89     dependent: clear(para_into_right).
% 1.68/1.89     dependent: set(para_from_vars).
% 1.68/1.89     dependent: set(eq_units_both_ways).
% 1.68/1.89     dependent: set(dynamic_demod_all).
% 1.68/1.89     dependent: set(dynamic_demod).
% 1.68/1.89     dependent: set(order_eq).
% 1.68/1.89     dependent: set(back_demod).
% 1.68/1.89     dependent: set(lrpo).
% 1.68/1.89     dependent: set(hyper_res).
% 1.68/1.89     dependent: set(unit_deletion).
% 1.68/1.89     dependent: set(factor).
% 1.68/1.89  
% 1.68/1.89  ------------> process usable:
% 1.68/1.89  ** KEPT (pick-wt=2): 1 [] -empty($c2).
% 1.68/1.89  ** KEPT (pick-wt=7): 3 [copy,2,flip.2] -subset($c4,singleton($c3))|empty_set!=$c4.
% 1.68/1.89  ** KEPT (pick-wt=8): 5 [copy,4,flip.2] -subset($c4,singleton($c3))|singleton($c3)!=$c4.
% 1.68/1.89  ** KEPT (pick-wt=11): 6 [] -subset(A,singleton(B))|A=empty_set|A=singleton(B).
% 1.68/1.89  ** KEPT (pick-wt=7): 7 [] subset(A,singleton(B))|A!=empty_set.
% 1.68/1.89  ** KEPT (pick-wt=8): 8 [] subset(A,singleton(B))|A!=singleton(B).
% 1.68/1.89  
% 1.68/1.89  ------------> process sos:
% 1.68/1.89  ** KEPT (pick-wt=3): 9 [] A=A.
% 1.68/1.89  ** KEPT (pick-wt=3): 10 [] subset(A,A).
% 1.68/1.89  ** KEPT (pick-wt=2): 11 [] empty(empty_set).
% 1.68/1.89  ** KEPT (pick-wt=2): 12 [] empty($c1).
% 1.68/1.89  ** KEPT (pick-wt=11): 14 [copy,13,flip.2,flip.3] subset($c4,singleton($c3))|empty_set=$c4|singleton($c3)=$c4.
% 1.68/1.89    Following clause subsumed by 9 during input processing: 0 [copy,9,flip.1] A=A.
% 1.68/1.89  
% 1.68/1.89  ======= end of input processing =======
% 1.68/1.89  
% 1.68/1.89  =========== start of search ===========
% 1.68/1.89  
% 1.68/1.89  -------- PROOF -------- 
% 1.68/1.89  
% 1.68/1.89  -----> EMPTY CLAUSE at   0.00 sec ----> 79 [back_demod,3,demod,69,unit_del,76,9] $F.
% 1.68/1.89  
% 1.68/1.89  Length of proof is 8.  Level of proof is 5.
% 1.68/1.89  
% 1.68/1.89  ---------------- PROOF ----------------
% 1.68/1.89  % SZS status Theorem
% 1.68/1.89  % SZS output start Refutation
% See solution above
% 1.68/1.89  ------------ end of proof -------------
% 1.68/1.89  
% 1.68/1.89  
% 1.68/1.89  Search stopped by max_proofs option.
% 1.68/1.89  
% 1.68/1.89  
% 1.68/1.89  Search stopped by max_proofs option.
% 1.68/1.89  
% 1.68/1.89  ============ end of search ============
% 1.68/1.89  
% 1.68/1.89  -------------- statistics -------------
% 1.68/1.89  clauses given                 15
% 1.68/1.89  clauses generated            130
% 1.68/1.89  clauses kept                  74
% 1.68/1.89  clauses forward subsumed      96
% 1.68/1.89  clauses back subsumed         21
% 1.68/1.89  Kbytes malloced              976
% 1.68/1.89  
% 1.68/1.89  ----------- times (seconds) -----------
% 1.68/1.89  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.68/1.89  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.68/1.89  wall-clock time        1             (0 hr, 0 min, 1 sec)
% 1.68/1.89  
% 1.68/1.89  That finishes the proof of the theorem.
% 1.68/1.89  
% 1.68/1.89  Process 7487 finished Wed Jul 27 07:42:38 2022
% 1.68/1.89  Otter interrupted
% 1.68/1.89  PROOF FOUND
%------------------------------------------------------------------------------