TSTP Solution File: SEU120+1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : SEU120+1 : TPTP v8.1.0. Released v3.3.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n012.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 13:14:48 EDT 2022

% Result   : Theorem 1.69s 1.93s
% Output   : Refutation 1.69s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    5
%            Number of leaves      :    8
% Syntax   : Number of clauses     :   15 (   7 unt;   2 nHn;  12 RR)
%            Number of literals    :   23 (   9 equ;   7 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    :    6 (   6 usr;   4 con; 0-2 aty)
%            Number of variables   :   11 (   2 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(2,axiom,
    ( A != empty_set
    | ~ in(B,A) ),
    file('SEU120+1.p',unknown),
    [] ).

cnf(3,axiom,
    ( ~ disjoint(A,B)
    | set_intersection2(A,B) = empty_set ),
    file('SEU120+1.p',unknown),
    [] ).

cnf(4,axiom,
    ( disjoint(A,B)
    | set_intersection2(A,B) != empty_set ),
    file('SEU120+1.p',unknown),
    [] ).

cnf(7,axiom,
    ( ~ disjoint(dollar_c5,dollar_c4)
    | in(dollar_c3,set_intersection2(dollar_c5,dollar_c4)) ),
    file('SEU120+1.p',unknown),
    [] ).

cnf(9,axiom,
    ( ~ in(A,set_intersection2(dollar_c5,dollar_c4))
    | disjoint(dollar_c5,dollar_c4) ),
    file('SEU120+1.p',unknown),
    [] ).

cnf(11,axiom,
    A = A,
    file('SEU120+1.p',unknown),
    [] ).

cnf(12,axiom,
    set_intersection2(A,B) = set_intersection2(B,A),
    file('SEU120+1.p',unknown),
    [] ).

cnf(13,axiom,
    ( A = empty_set
    | in(dollar_f1(A),A) ),
    file('SEU120+1.p',unknown),
    [] ).

cnf(26,plain,
    ( ~ disjoint(dollar_c5,dollar_c4)
    | in(dollar_c3,set_intersection2(dollar_c4,dollar_c5)) ),
    inference(para_from,[status(thm),theory(equality)],[12,7]),
    [iquote('para_from,12.1.1,7.2.2')] ).

cnf(53,plain,
    ( set_intersection2(dollar_c5,dollar_c4) = empty_set
    | disjoint(dollar_c5,dollar_c4) ),
    inference(hyper,[status(thm)],[13,9]),
    [iquote('hyper,13,9')] ).

cnf(180,plain,
    set_intersection2(dollar_c5,dollar_c4) = empty_set,
    inference(factor_simp,[status(thm)],[inference(hyper,[status(thm)],[53,3])]),
    [iquote('hyper,53,3,factor_simp')] ).

cnf(212,plain,
    disjoint(dollar_c5,dollar_c4),
    inference(hyper,[status(thm)],[180,4]),
    [iquote('hyper,180,4')] ).

cnf(218,plain,
    set_intersection2(dollar_c4,dollar_c5) = empty_set,
    inference(para_into,[status(thm),theory(equality)],[180,12]),
    [iquote('para_into,180.1.1,12.1.1')] ).

cnf(219,plain,
    in(dollar_c3,empty_set),
    inference(unit_del,[status(thm)],[inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[26]),218]),212]),
    [iquote('back_demod,26,demod,218,unit_del,212')] ).

cnf(252,plain,
    $false,
    inference(hyper,[status(thm)],[219,2,11]),
    [iquote('hyper,219,2,11')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.11  % Problem  : SEU120+1 : TPTP v8.1.0. Released v3.3.0.
% 0.11/0.12  % Command  : otter-tptp-script %s
% 0.12/0.33  % Computer : n012.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 07:52:50 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 n012.cluster.edu,
% 1.69/1.93  Wed Jul 27 07:52:50 2022
% 1.69/1.93  The command was "./otter".  The process ID is 13543.
% 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 (set_intersection2(A,B)=set_intersection2(B,A)).
% 1.69/1.93  all A (A=empty_set<-> (all B (-in(B,A)))).
% 1.69/1.93  all A B (disjoint(A,B)<->set_intersection2(A,B)=empty_set).
% 1.69/1.93  $T.
% 1.69/1.93  $T.
% 1.69/1.93  empty(empty_set).
% 1.69/1.93  all A B (set_intersection2(A,A)=A).
% 1.69/1.93  exists A empty(A).
% 1.69/1.93  exists A (-empty(A)).
% 1.69/1.93  all A B (disjoint(A,B)->disjoint(B,A)).
% 1.69/1.93  -(all A B (-(-disjoint(A,B)& (all C (-in(C,set_intersection2(A,B)))))& -((exists C in(C,set_intersection2(A,B)))&disjoint(A,B)))).
% 1.69/1.93  end_of_list.
% 1.69/1.93  
% 1.69/1.93  -------> usable clausifies to:
% 1.69/1.93  
% 1.69/1.93  list(usable).
% 1.69/1.93  0 [] A=A.
% 1.69/1.93  0 [] -in(A,B)| -in(B,A).
% 1.69/1.93  0 [] set_intersection2(A,B)=set_intersection2(B,A).
% 1.69/1.93  0 [] A!=empty_set| -in(B,A).
% 1.69/1.93  0 [] A=empty_set|in($f1(A),A).
% 1.69/1.93  0 [] -disjoint(A,B)|set_intersection2(A,B)=empty_set.
% 1.69/1.93  0 [] disjoint(A,B)|set_intersection2(A,B)!=empty_set.
% 1.69/1.93  0 [] $T.
% 1.69/1.93  0 [] $T.
% 1.69/1.93  0 [] empty(empty_set).
% 1.69/1.93  0 [] set_intersection2(A,A)=A.
% 1.69/1.93  0 [] empty($c1).
% 1.69/1.93  0 [] -empty($c2).
% 1.69/1.93  0 [] -disjoint(A,B)|disjoint(B,A).
% 1.69/1.93  0 [] -disjoint($c5,$c4)|in($c3,set_intersection2($c5,$c4)).
% 1.69/1.93  0 [] -in(C,set_intersection2($c5,$c4))|in($c3,set_intersection2($c5,$c4)).
% 1.69/1.93  0 [] -in(C,set_intersection2($c5,$c4))|disjoint($c5,$c4).
% 1.69/1.93  end_of_list.
% 1.69/1.93  
% 1.69/1.93  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=2.
% 1.69/1.93  
% 1.69/1.93  This ia a non-Horn set with equality.  The strategy will be
% 1.69/1.93  Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.69/1.93  deletion, with positive clauses in sos and nonpositive
% 1.69/1.93  clauses in usable.
% 1.69/1.93  
% 1.69/1.93     dependent: set(knuth_bendix).
% 1.69/1.93     dependent: set(anl_eq).
% 1.69/1.93     dependent: set(para_from).
% 1.69/1.93     dependent: set(para_into).
% 1.69/1.93     dependent: clear(para_from_right).
% 1.69/1.93     dependent: clear(para_into_right).
% 1.69/1.93     dependent: set(para_from_vars).
% 1.69/1.93     dependent: set(eq_units_both_ways).
% 1.69/1.93     dependent: set(dynamic_demod_all).
% 1.69/1.93     dependent: set(dynamic_demod).
% 1.69/1.93     dependent: set(order_eq).
% 1.69/1.93     dependent: set(back_demod).
% 1.69/1.93     dependent: set(lrpo).
% 1.69/1.93     dependent: set(hyper_res).
% 1.69/1.93     dependent: set(unit_deletion).
% 1.69/1.93     dependent: set(factor).
% 1.69/1.93  
% 1.69/1.93  ------------> process usable:
% 1.69/1.93  ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.69/1.93  ** KEPT (pick-wt=6): 2 [] A!=empty_set| -in(B,A).
% 1.69/1.93  ** KEPT (pick-wt=8): 3 [] -disjoint(A,B)|set_intersection2(A,B)=empty_set.
% 1.69/1.93  ** KEPT (pick-wt=8): 4 [] disjoint(A,B)|set_intersection2(A,B)!=empty_set.
% 1.69/1.93  ** KEPT (pick-wt=2): 5 [] -empty($c2).
% 1.69/1.93  ** KEPT (pick-wt=6): 6 [] -disjoint(A,B)|disjoint(B,A).
% 1.69/1.93  ** KEPT (pick-wt=8): 7 [] -disjoint($c5,$c4)|in($c3,set_intersection2($c5,$c4)).
% 1.69/1.93  ** KEPT (pick-wt=10): 8 [] -in(A,set_intersection2($c5,$c4))|in($c3,set_intersection2($c5,$c4)).
% 1.69/1.93  ** KEPT (pick-wt=8): 9 [] -in(A,set_intersection2($c5,$c4))|disjoint($c5,$c4).
% 1.69/1.93  
% 1.69/1.93  ------------> process sos:
% 1.69/1.93  ** KEPT (pick-wt=3): 11 [] A=A.
% 1.69/1.93  ** KEPT (pick-wt=7): 12 [] set_intersection2(A,B)=set_intersection2(B,A).
% 1.69/1.93  ** KEPT (pick-wt=7): 13 [] A=empty_set|in($f1(A),A).
% 1.69/1.93  ** KEPT (pick-wt=2): 14 [] empty(empty_set).
% 1.69/1.93  ** KEPT (pick-wt=5): 15 [] set_intersection2(A,A)=A.
% 1.69/1.93  ---> New Demodulator: 16 [new_demod,15] set_intersection2(A,A)=A.
% 1.69/1.93  ** KEPT (pick-wt=2): 17 [] empty($c1).
% 1.69/1.93    Following clause subsumed by 11 during input processing: 0 [copy,11,flip.1] A=A.
% 1.69/1.93    Following clause subsumed by 12 during input processing: 0 [copy,12,flip.1] set_intersection2(A,B)=set_intersection2(B,A).
% 1.69/1.93  >>>> Starting back demodulation with 16.
% 1.69/1.93  
% 1.69/1.93  ======= end of input processing =======
% 1.69/1.93  
% 1.69/1.93  =========== start of search ===========
% 1.69/1.93  
% 1.69/1.93  -------- PROOF -------- 
% 1.69/1.93  
% 1.69/1.93  -----> EMPTY CLAUSE at   0.01 sec ----> 252 [hyper,219,2,11] $F.
% 1.69/1.93  
% 1.69/1.93  Length of proof is 6.  Level of proof is 4.
% 1.69/1.93  
% 1.69/1.93  ---------------- PROOF ----------------
% 1.69/1.93  % SZS status Theorem
% 1.69/1.93  % SZS output start Refutation
% See solution above
% 1.69/1.93  ------------ end of proof -------------
% 1.69/1.93  
% 1.69/1.93  
% 1.69/1.93  Search stopped by max_proofs option.
% 1.69/1.93  
% 1.69/1.93  
% 1.69/1.93  Search stopped by max_proofs option.
% 1.69/1.93  
% 1.69/1.93  ============ end of search ============
% 1.69/1.93  
% 1.69/1.93  -------------- statistics -------------
% 1.69/1.93  clauses given                 34
% 1.69/1.93  clauses generated            382
% 1.69/1.93  clauses kept                 248
% 1.69/1.93  clauses forward subsumed     218
% 1.69/1.93  clauses back subsumed         66
% 1.69/1.93  Kbytes malloced              976
% 1.69/1.93  
% 1.69/1.93  ----------- times (seconds) -----------
% 1.69/1.93  user CPU time          0.01          (0 hr, 0 min, 0 sec)
% 1.69/1.93  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.69/1.93  wall-clock time        2             (0 hr, 0 min, 2 sec)
% 1.69/1.93  
% 1.69/1.93  That finishes the proof of the theorem.
% 1.69/1.93  
% 1.69/1.93  Process 13543 finished Wed Jul 27 07:52:52 2022
% 1.69/1.93  Otter interrupted
% 1.69/1.93  PROOF FOUND
%------------------------------------------------------------------------------