TSTP Solution File: SET926+1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : SET926+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n008.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 13:14:32 EDT 2022

% Result   : Theorem 1.69s 1.90s
% Output   : Refutation 1.69s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    3
%            Number of leaves      :    5
% Syntax   : Number of clauses     :    8 (   6 unt;   1 nHn;   6 RR)
%            Number of literals    :   10 (   6 equ;   3 neg)
%            Maximal clause size   :    2 (   1 avg)
%            Maximal term depth    :    3 (   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   :    5 (   0 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(4,axiom,
    ( set_difference(singleton(A),B) = empty_set
    | ~ in(A,B) ),
    file('SET926+1.p',unknown),
    [] ).

cnf(6,axiom,
    set_difference(singleton(dollar_c4),dollar_c3) != empty_set,
    file('SET926+1.p',unknown),
    [] ).

cnf(7,axiom,
    set_difference(singleton(dollar_c4),dollar_c3) != singleton(dollar_c4),
    file('SET926+1.p',unknown),
    [] ).

cnf(9,axiom,
    A = A,
    file('SET926+1.p',unknown),
    [] ).

cnf(11,axiom,
    ( set_difference(singleton(A),B) = singleton(A)
    | in(A,B) ),
    file('SET926+1.p',unknown),
    [] ).

cnf(17,plain,
    in(dollar_c4,dollar_c3),
    inference(unit_del,[status(thm)],[inference(para_from,[status(thm),theory(equality)],[11,7]),9]),
    [iquote('para_from,11.1.1,7.1.1,unit_del,9')] ).

cnf(19,plain,
    set_difference(singleton(dollar_c4),dollar_c3) = empty_set,
    inference(hyper,[status(thm)],[17,4]),
    [iquote('hyper,17,4')] ).

cnf(21,plain,
    $false,
    inference(binary,[status(thm)],[19,6]),
    [iquote('binary,19.1,6.1')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : SET926+1 : TPTP v8.1.0. Released v3.2.0.
% 0.12/0.12  % Command  : otter-tptp-script %s
% 0.12/0.33  % Computer : n008.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 10:42:08 EDT 2022
% 0.12/0.33  % CPUTime  : 
% 1.69/1.90  ----- Otter 3.3f, August 2004 -----
% 1.69/1.90  The process was started by sandbox on n008.cluster.edu,
% 1.69/1.90  Wed Jul 27 10:42:08 2022
% 1.69/1.90  The command was "./otter".  The process ID is 22256.
% 1.69/1.90  
% 1.69/1.90  set(prolog_style_variables).
% 1.69/1.90  set(auto).
% 1.69/1.90     dependent: set(auto1).
% 1.69/1.90     dependent: set(process_input).
% 1.69/1.90     dependent: clear(print_kept).
% 1.69/1.90     dependent: clear(print_new_demod).
% 1.69/1.90     dependent: clear(print_back_demod).
% 1.69/1.90     dependent: clear(print_back_sub).
% 1.69/1.90     dependent: set(control_memory).
% 1.69/1.90     dependent: assign(max_mem, 12000).
% 1.69/1.90     dependent: assign(pick_given_ratio, 4).
% 1.69/1.90     dependent: assign(stats_level, 1).
% 1.69/1.90     dependent: assign(max_seconds, 10800).
% 1.69/1.90  clear(print_given).
% 1.69/1.90  
% 1.69/1.90  formula_list(usable).
% 1.69/1.90  all A (A=A).
% 1.69/1.90  all A B (in(A,B)-> -in(B,A)).
% 1.69/1.90  empty(empty_set).
% 1.69/1.90  all A B (set_difference(singleton(A),B)=singleton(A)<-> -in(A,B)).
% 1.69/1.90  all A B (set_difference(singleton(A),B)=empty_set<->in(A,B)).
% 1.69/1.90  exists A empty(A).
% 1.69/1.90  exists A (-empty(A)).
% 1.69/1.90  -(all A B (set_difference(singleton(A),B)=empty_set|set_difference(singleton(A),B)=singleton(A))).
% 1.69/1.90  end_of_list.
% 1.69/1.90  
% 1.69/1.90  -------> usable clausifies to:
% 1.69/1.90  
% 1.69/1.90  list(usable).
% 1.69/1.90  0 [] A=A.
% 1.69/1.90  0 [] -in(A,B)| -in(B,A).
% 1.69/1.90  0 [] empty(empty_set).
% 1.69/1.90  0 [] set_difference(singleton(A),B)!=singleton(A)| -in(A,B).
% 1.69/1.90  0 [] set_difference(singleton(A),B)=singleton(A)|in(A,B).
% 1.69/1.90  0 [] set_difference(singleton(A),B)!=empty_set|in(A,B).
% 1.69/1.90  0 [] set_difference(singleton(A),B)=empty_set| -in(A,B).
% 1.69/1.90  0 [] empty($c1).
% 1.69/1.90  0 [] -empty($c2).
% 1.69/1.90  0 [] set_difference(singleton($c4),$c3)!=empty_set.
% 1.69/1.90  0 [] set_difference(singleton($c4),$c3)!=singleton($c4).
% 1.69/1.90  end_of_list.
% 1.69/1.90  
% 1.69/1.90  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=2.
% 1.69/1.90  
% 1.69/1.90  This ia a non-Horn set with equality.  The strategy will be
% 1.69/1.90  Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.69/1.90  deletion, with positive clauses in sos and nonpositive
% 1.69/1.90  clauses in usable.
% 1.69/1.90  
% 1.69/1.90     dependent: set(knuth_bendix).
% 1.69/1.90     dependent: set(anl_eq).
% 1.69/1.90     dependent: set(para_from).
% 1.69/1.90     dependent: set(para_into).
% 1.69/1.90     dependent: clear(para_from_right).
% 1.69/1.90     dependent: clear(para_into_right).
% 1.69/1.90     dependent: set(para_from_vars).
% 1.69/1.90     dependent: set(eq_units_both_ways).
% 1.69/1.90     dependent: set(dynamic_demod_all).
% 1.69/1.90     dependent: set(dynamic_demod).
% 1.69/1.90     dependent: set(order_eq).
% 1.69/1.90     dependent: set(back_demod).
% 1.69/1.90     dependent: set(lrpo).
% 1.69/1.90     dependent: set(hyper_res).
% 1.69/1.90     dependent: set(unit_deletion).
% 1.69/1.90     dependent: set(factor).
% 1.69/1.90  
% 1.69/1.90  ------------> process usable:
% 1.69/1.90  ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.69/1.90  ** KEPT (pick-wt=10): 2 [] set_difference(singleton(A),B)!=singleton(A)| -in(A,B).
% 1.69/1.90  ** KEPT (pick-wt=9): 3 [] set_difference(singleton(A),B)!=empty_set|in(A,B).
% 1.69/1.90  ** KEPT (pick-wt=9): 4 [] set_difference(singleton(A),B)=empty_set| -in(A,B).
% 1.69/1.90  ** KEPT (pick-wt=2): 5 [] -empty($c2).
% 1.69/1.90  ** KEPT (pick-wt=6): 6 [] set_difference(singleton($c4),$c3)!=empty_set.
% 1.69/1.90  ** KEPT (pick-wt=7): 7 [] set_difference(singleton($c4),$c3)!=singleton($c4).
% 1.69/1.90  
% 1.69/1.90  ------------> process sos:
% 1.69/1.90  ** KEPT (pick-wt=3): 9 [] A=A.
% 1.69/1.90  ** KEPT (pick-wt=2): 10 [] empty(empty_set).
% 1.69/1.90  ** KEPT (pick-wt=10): 11 [] set_difference(singleton(A),B)=singleton(A)|in(A,B).
% 1.69/1.90  ** KEPT (pick-wt=2): 12 [] empty($c1).
% 1.69/1.90    Following clause subsumed by 9 during input processing: 0 [copy,9,flip.1] A=A.
% 1.69/1.90  
% 1.69/1.90  ======= end of input processing =======
% 1.69/1.90  
% 1.69/1.90  =========== start of search ===========
% 1.69/1.90  
% 1.69/1.90  -------- PROOF -------- 
% 1.69/1.90  
% 1.69/1.90  ----> UNIT CONFLICT at   0.00 sec ----> 21 [binary,19.1,6.1] $F.
% 1.69/1.90  
% 1.69/1.90  Length of proof is 2.  Level of proof is 2.
% 1.69/1.90  
% 1.69/1.90  ---------------- PROOF ----------------
% 1.69/1.90  % SZS status Theorem
% 1.69/1.90  % SZS output start Refutation
% See solution above
% 1.69/1.90  ------------ end of proof -------------
% 1.69/1.90  
% 1.69/1.90  
% 1.69/1.90  Search stopped by max_proofs option.
% 1.69/1.90  
% 1.69/1.90  
% 1.69/1.90  Search stopped by max_proofs option.
% 1.69/1.90  
% 1.69/1.90  ============ end of search ============
% 1.69/1.90  
% 1.69/1.90  -------------- statistics -------------
% 1.69/1.90  clauses given                  5
% 1.69/1.90  clauses generated             15
% 1.69/1.90  clauses kept                  18
% 1.69/1.90  clauses forward subsumed       6
% 1.69/1.90  clauses back subsumed          0
% 1.69/1.90  Kbytes malloced              976
% 1.69/1.90  
% 1.69/1.90  ----------- times (seconds) -----------
% 1.69/1.90  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.69/1.90  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.69/1.90  wall-clock time        2             (0 hr, 0 min, 2 sec)
% 1.69/1.90  
% 1.69/1.90  That finishes the proof of the theorem.
% 1.69/1.90  
% 1.69/1.90  Process 22256 finished Wed Jul 27 10:42:10 2022
% 1.69/1.90  Otter interrupted
% 1.69/1.90  PROOF FOUND
%------------------------------------------------------------------------------