TSTP Solution File: SEU134+1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : SEU134+1 : TPTP v8.1.0. Released v3.3.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n024.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 13:14:51 EDT 2022

% Result   : Theorem 1.68s 1.89s
% Output   : Refutation 1.68s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    5
%            Number of leaves      :    7
% Syntax   : Number of clauses     :   16 (   7 unt;   2 nHn;  15 RR)
%            Number of literals    :   25 (  12 equ;  10 neg)
%            Maximal clause size   :    2 (   1 avg)
%            Maximal term depth    :    2 (   1 avg)
%            Number of predicates  :    4 (   2 usr;   1 prp; 0-2 aty)
%            Number of functors    :    5 (   5 usr;   4 con; 0-2 aty)
%            Number of variables   :   10 (   0 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(5,axiom,
    ( ~ empty(A)
    | A = empty_set ),
    file('SEU134+1.p',unknown),
    [] ).

cnf(6,axiom,
    ( set_difference(dollar_c4,dollar_c3) != empty_set
    | ~ subset(dollar_c4,dollar_c3) ),
    file('SEU134+1.p',unknown),
    [] ).

cnf(7,axiom,
    ( set_difference(A,B) != empty_set
    | subset(A,B) ),
    file('SEU134+1.p',unknown),
    [] ).

cnf(8,axiom,
    ( set_difference(A,B) = empty_set
    | ~ subset(A,B) ),
    file('SEU134+1.p',unknown),
    [] ).

cnf(11,axiom,
    A = A,
    file('SEU134+1.p',unknown),
    [] ).

cnf(12,axiom,
    empty(dollar_c1),
    file('SEU134+1.p',unknown),
    [] ).

cnf(19,axiom,
    ( set_difference(dollar_c4,dollar_c3) = empty_set
    | subset(dollar_c4,dollar_c3) ),
    file('SEU134+1.p',unknown),
    [] ).

cnf(21,plain,
    empty_set = dollar_c1,
    inference(flip,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[12,5])]),
    [iquote('hyper,12,5,flip.1')] ).

cnf(22,plain,
    ( set_difference(dollar_c4,dollar_c3) = dollar_c1
    | subset(dollar_c4,dollar_c3) ),
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[19]),21]),
    [iquote('back_demod,19,demod,21')] ).

cnf(27,plain,
    ( set_difference(A,B) = dollar_c1
    | ~ subset(A,B) ),
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[8]),21]),
    [iquote('back_demod,8,demod,21')] ).

cnf(28,plain,
    ( set_difference(A,B) != dollar_c1
    | subset(A,B) ),
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[7]),21]),
    [iquote('back_demod,7,demod,21')] ).

cnf(29,plain,
    ( set_difference(dollar_c4,dollar_c3) != dollar_c1
    | ~ subset(dollar_c4,dollar_c3) ),
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[6]),21]),
    [iquote('back_demod,6,demod,21')] ).

cnf(39,plain,
    set_difference(dollar_c4,dollar_c3) = dollar_c1,
    inference(factor_simp,[status(thm)],[inference(hyper,[status(thm)],[27,22])]),
    [iquote('hyper,27,22,factor_simp')] ).

cnf(42,plain,
    ~ subset(dollar_c4,dollar_c3),
    inference(unit_del,[status(thm)],[inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[29]),39]),11]),
    [iquote('back_demod,29,demod,39,unit_del,11')] ).

cnf(52,plain,
    subset(dollar_c4,dollar_c3),
    inference(hyper,[status(thm)],[28,39]),
    [iquote('hyper,28,38')] ).

cnf(53,plain,
    $false,
    inference(binary,[status(thm)],[52,42]),
    [iquote('binary,52.1,42.1')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12  % Problem  : SEU134+1 : TPTP v8.1.0. Released v3.3.0.
% 0.03/0.12  % Command  : otter-tptp-script %s
% 0.12/0.33  % Computer : n024.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 08:05:40 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 sandbox2 on n024.cluster.edu,
% 1.68/1.89  Wed Jul 27 08:05:40 2022
% 1.68/1.89  The command was "./otter".  The process ID is 29051.
% 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 (in(A,B)-> -in(B,A)).
% 1.68/1.89  exists A empty(A).
% 1.68/1.89  exists A (-empty(A)).
% 1.68/1.89  all A B (-(in(A,B)&empty(B))).
% 1.68/1.89  all A B (-(empty(A)&A!=B&empty(B))).
% 1.68/1.89  all A B subset(A,A).
% 1.68/1.89  $T.
% 1.68/1.89  $T.
% 1.68/1.89  empty(empty_set).
% 1.68/1.89  all A (set_difference(A,empty_set)=A).
% 1.68/1.89  all A (set_difference(empty_set,A)=empty_set).
% 1.68/1.89  all A (empty(A)->A=empty_set).
% 1.68/1.89  -(all A B (set_difference(A,B)=empty_set<->subset(A,B))).
% 1.68/1.89  all A B (set_difference(A,B)=empty_set<->subset(A,B)).
% 1.68/1.89  end_of_list.
% 1.68/1.89  
% 1.68/1.89  -------> usable clausifies to:
% 1.68/1.89  
% 1.68/1.89  list(usable).
% 1.68/1.89  0 [] A=A.
% 1.68/1.89  0 [] -in(A,B)| -in(B,A).
% 1.68/1.89  0 [] empty($c1).
% 1.68/1.89  0 [] -empty($c2).
% 1.68/1.89  0 [] -in(A,B)| -empty(B).
% 1.68/1.89  0 [] -empty(A)|A=B| -empty(B).
% 1.68/1.89  0 [] subset(A,A).
% 1.68/1.89  0 [] $T.
% 1.68/1.89  0 [] $T.
% 1.68/1.89  0 [] empty(empty_set).
% 1.68/1.89  0 [] set_difference(A,empty_set)=A.
% 1.68/1.89  0 [] set_difference(empty_set,A)=empty_set.
% 1.68/1.89  0 [] -empty(A)|A=empty_set.
% 1.68/1.89  0 [] set_difference($c4,$c3)=empty_set|subset($c4,$c3).
% 1.68/1.89  0 [] set_difference($c4,$c3)!=empty_set| -subset($c4,$c3).
% 1.68/1.89  0 [] set_difference(A,B)!=empty_set|subset(A,B).
% 1.68/1.89  0 [] set_difference(A,B)=empty_set| -subset(A,B).
% 1.68/1.89  end_of_list.
% 1.68/1.89  
% 1.68/1.89  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.68/1.89  
% 1.68/1.89  This ia a non-Horn set with equality.  The strategy will be
% 1.68/1.89  Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.68/1.89  deletion, with positive clauses in sos and nonpositive
% 1.68/1.89  clauses in usable.
% 1.68/1.89  
% 1.68/1.89     dependent: set(knuth_bendix).
% 1.68/1.89     dependent: set(anl_eq).
% 1.68/1.89     dependent: set(para_from).
% 1.68/1.89     dependent: set(para_into).
% 1.68/1.89     dependent: clear(para_from_right).
% 1.68/1.89     dependent: clear(para_into_right).
% 1.68/1.89     dependent: set(para_from_vars).
% 1.68/1.89     dependent: set(eq_units_both_ways).
% 1.68/1.89     dependent: set(dynamic_demod_all).
% 1.68/1.89     dependent: set(dynamic_demod).
% 1.68/1.89     dependent: set(order_eq).
% 1.68/1.89     dependent: set(back_demod).
% 1.68/1.89     dependent: set(lrpo).
% 1.68/1.89     dependent: set(hyper_res).
% 1.68/1.89     dependent: set(unit_deletion).
% 1.68/1.89     dependent: set(factor).
% 1.68/1.89  
% 1.68/1.89  ------------> process usable:
% 1.68/1.89  ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.68/1.89  ** KEPT (pick-wt=2): 2 [] -empty($c2).
% 1.68/1.89  ** KEPT (pick-wt=5): 3 [] -in(A,B)| -empty(B).
% 1.68/1.89  ** KEPT (pick-wt=7): 4 [] -empty(A)|A=B| -empty(B).
% 1.68/1.89  ** KEPT (pick-wt=5): 5 [] -empty(A)|A=empty_set.
% 1.68/1.89  ** KEPT (pick-wt=8): 6 [] set_difference($c4,$c3)!=empty_set| -subset($c4,$c3).
% 1.68/1.89  ** KEPT (pick-wt=8): 7 [] set_difference(A,B)!=empty_set|subset(A,B).
% 1.68/1.89  ** KEPT (pick-wt=8): 8 [] set_difference(A,B)=empty_set| -subset(A,B).
% 1.68/1.89  
% 1.68/1.89  ------------> process sos:
% 1.68/1.89  ** KEPT (pick-wt=3): 11 [] A=A.
% 1.68/1.89  ** KEPT (pick-wt=2): 12 [] empty($c1).
% 1.68/1.89  ** KEPT (pick-wt=3): 13 [] subset(A,A).
% 1.68/1.89  ** KEPT (pick-wt=2): 14 [] empty(empty_set).
% 1.68/1.89  ** KEPT (pick-wt=5): 15 [] set_difference(A,empty_set)=A.
% 1.68/1.89  ---> New Demodulator: 16 [new_demod,15] set_difference(A,empty_set)=A.
% 1.68/1.89  ** KEPT (pick-wt=5): 17 [] set_difference(empty_set,A)=empty_set.
% 1.68/1.89  ---> New Demodulator: 18 [new_demod,17] set_difference(empty_set,A)=empty_set.
% 1.68/1.89  ** KEPT (pick-wt=8): 19 [] set_difference($c4,$c3)=empty_set|subset($c4,$c3).
% 1.68/1.89    Following clause subsumed by 11 during input processing: 0 [copy,11,flip.1] A=A.
% 1.68/1.89  11 back subsumes 10.
% 1.68/1.89  >>>> Starting back demodulation with 16.
% 1.68/1.89  >>>> Starting back demodulation with 18.
% 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  ----> UNIT CONFLICT at   0.00 sec ----> 53 [binary,52.1,42.1] $F.
% 1.68/1.89  
% 1.68/1.89  Length of proof is 8.  Level of proof is 4.
% 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                 16
% 1.68/1.89  clauses generated            112
% 1.68/1.89  clauses kept                  45
% 1.68/1.89  clauses forward subsumed      93
% 1.68/1.89  clauses back subsumed          3
% 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 29051 finished Wed Jul 27 08:05:41 2022
% 1.68/1.89  Otter interrupted
% 1.68/1.89  PROOF FOUND
%------------------------------------------------------------------------------