TSTP Solution File: ROB021-1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : ROB021-1 : TPTP v8.1.0. Released v1.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n009.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 13:12:23 EDT 2022

% Result   : Unsatisfiable 1.75s 1.92s
% Output   : Refutation 1.75s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    3
%            Number of leaves      :    5
% Syntax   : Number of clauses     :    9 (   8 unt;   0 nHn;   5 RR)
%            Number of literals    :   10 (   9 equ;   4 neg)
%            Maximal clause size   :    2 (   1 avg)
%            Maximal term depth    :    6 (   2 avg)
%            Number of predicates  :    2 (   0 usr;   1 prp; 0-2 aty)
%            Number of functors    :    4 (   4 usr;   2 con; 0-2 aty)
%            Number of variables   :    9 (   0 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(1,axiom,
    ( negate(A) != negate(B)
    | A = B ),
    file('ROB021-1.p',unknown),
    [] ).

cnf(2,axiom,
    add(negate(add(a,negate(b))),negate(add(negate(a),negate(b)))) != b,
    file('ROB021-1.p',unknown),
    [] ).

cnf(3,axiom,
    A = A,
    file('ROB021-1.p',unknown),
    [] ).

cnf(4,axiom,
    add(A,B) = add(B,A),
    file('ROB021-1.p',unknown),
    [] ).

cnf(7,axiom,
    negate(add(negate(add(A,B)),negate(add(A,negate(B))))) = A,
    file('ROB021-1.p',unknown),
    [] ).

cnf(11,plain,
    add(negate(add(a,negate(b))),negate(add(negate(b),negate(a)))) != b,
    inference(para_from,[status(thm),theory(equality)],[4,2]),
    [iquote('para_from,4.1.1,2.1.1.2.1')] ).

cnf(33,plain,
    add(negate(add(negate(A),B)),negate(add(negate(A),negate(B)))) = A,
    inference(hyper,[status(thm)],[7,1]),
    [iquote('hyper,7,1')] ).

cnf(466,plain,
    b != b,
    inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[11,4]),33]),
    [iquote('para_into,11.1.1.1.1,4.1.1,demod,33')] ).

cnf(467,plain,
    $false,
    inference(binary,[status(thm)],[466,3]),
    [iquote('binary,466.1,3.1')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12  % Problem  : ROB021-1 : TPTP v8.1.0. Released v1.0.0.
% 0.11/0.13  % Command  : otter-tptp-script %s
% 0.13/0.34  % Computer : n009.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit : 300
% 0.13/0.34  % WCLimit  : 300
% 0.13/0.34  % DateTime : Wed Jul 27 04:01:21 EDT 2022
% 0.13/0.34  % CPUTime  : 
% 1.75/1.92  ----- Otter 3.3f, August 2004 -----
% 1.75/1.92  The process was started by sandbox2 on n009.cluster.edu,
% 1.75/1.92  Wed Jul 27 04:01:21 2022
% 1.75/1.92  The command was "./otter".  The process ID is 3662.
% 1.75/1.92  
% 1.75/1.92  set(prolog_style_variables).
% 1.75/1.92  set(auto).
% 1.75/1.92     dependent: set(auto1).
% 1.75/1.92     dependent: set(process_input).
% 1.75/1.92     dependent: clear(print_kept).
% 1.75/1.92     dependent: clear(print_new_demod).
% 1.75/1.92     dependent: clear(print_back_demod).
% 1.75/1.92     dependent: clear(print_back_sub).
% 1.75/1.92     dependent: set(control_memory).
% 1.75/1.92     dependent: assign(max_mem, 12000).
% 1.75/1.92     dependent: assign(pick_given_ratio, 4).
% 1.75/1.92     dependent: assign(stats_level, 1).
% 1.75/1.92     dependent: assign(max_seconds, 10800).
% 1.75/1.92  clear(print_given).
% 1.75/1.92  
% 1.75/1.92  list(usable).
% 1.75/1.92  0 [] A=A.
% 1.75/1.92  0 [] add(X,Y)=add(Y,X).
% 1.75/1.92  0 [] add(add(X,Y),Z)=add(X,add(Y,Z)).
% 1.75/1.92  0 [] negate(add(negate(add(X,Y)),negate(add(X,negate(Y)))))=X.
% 1.75/1.92  0 [] negate(X)!=negate(Y)|X=Y.
% 1.75/1.92  0 [] add(negate(add(a,negate(b))),negate(add(negate(a),negate(b))))!=b.
% 1.75/1.92  end_of_list.
% 1.75/1.92  
% 1.75/1.92  SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=2.
% 1.75/1.92  
% 1.75/1.92  This is a Horn set with equality.  The strategy will be
% 1.75/1.92  Knuth-Bendix and hyper_res, with positive clauses in
% 1.75/1.92  sos and nonpositive clauses in usable.
% 1.75/1.92  
% 1.75/1.92     dependent: set(knuth_bendix).
% 1.75/1.92     dependent: set(anl_eq).
% 1.75/1.92     dependent: set(para_from).
% 1.75/1.92     dependent: set(para_into).
% 1.75/1.92     dependent: clear(para_from_right).
% 1.75/1.92     dependent: clear(para_into_right).
% 1.75/1.92     dependent: set(para_from_vars).
% 1.75/1.92     dependent: set(eq_units_both_ways).
% 1.75/1.92     dependent: set(dynamic_demod_all).
% 1.75/1.92     dependent: set(dynamic_demod).
% 1.75/1.92     dependent: set(order_eq).
% 1.75/1.92     dependent: set(back_demod).
% 1.75/1.92     dependent: set(lrpo).
% 1.75/1.92     dependent: set(hyper_res).
% 1.75/1.92     dependent: clear(order_hyper).
% 1.75/1.92  
% 1.75/1.92  ------------> process usable:
% 1.75/1.92  ** KEPT (pick-wt=8): 1 [] negate(A)!=negate(B)|A=B.
% 1.75/1.92  ** KEPT (pick-wt=14): 2 [] add(negate(add(a,negate(b))),negate(add(negate(a),negate(b))))!=b.
% 1.75/1.92  
% 1.75/1.92  ------------> process sos:
% 1.75/1.92  ** KEPT (pick-wt=3): 3 [] A=A.
% 1.75/1.92  ** KEPT (pick-wt=7): 4 [] add(A,B)=add(B,A).
% 1.75/1.92  ** KEPT (pick-wt=11): 5 [] add(add(A,B),C)=add(A,add(B,C)).
% 1.75/1.92  ---> New Demodulator: 6 [new_demod,5] add(add(A,B),C)=add(A,add(B,C)).
% 1.75/1.92  ** KEPT (pick-wt=13): 7 [] negate(add(negate(add(A,B)),negate(add(A,negate(B)))))=A.
% 1.75/1.92  ---> New Demodulator: 8 [new_demod,7] negate(add(negate(add(A,B)),negate(add(A,negate(B)))))=A.
% 1.75/1.92    Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.75/1.92    Following clause subsumed by 4 during input processing: 0 [copy,4,flip.1] add(A,B)=add(B,A).
% 1.75/1.92  >>>> Starting back demodulation with 6.
% 1.75/1.92  >>>> Starting back demodulation with 8.
% 1.75/1.92  
% 1.75/1.92  ======= end of input processing =======
% 1.75/1.92  
% 1.75/1.92  =========== start of search ===========
% 1.75/1.92  
% 1.75/1.92  -------- PROOF -------- 
% 1.75/1.92  
% 1.75/1.92  ----> UNIT CONFLICT at   0.03 sec ----> 467 [binary,466.1,3.1] $F.
% 1.75/1.92  
% 1.75/1.92  Length of proof is 3.  Level of proof is 2.
% 1.75/1.92  
% 1.75/1.92  ---------------- PROOF ----------------
% 1.75/1.92  % SZS status Unsatisfiable
% 1.75/1.92  % SZS output start Refutation
% See solution above
% 1.75/1.92  ------------ end of proof -------------
% 1.75/1.92  
% 1.75/1.92  
% 1.75/1.92  Search stopped by max_proofs option.
% 1.75/1.92  
% 1.75/1.92  
% 1.75/1.92  Search stopped by max_proofs option.
% 1.75/1.92  
% 1.75/1.92  ============ end of search ============
% 1.75/1.92  
% 1.75/1.92  -------------- statistics -------------
% 1.75/1.92  clauses given                 21
% 1.75/1.92  clauses generated            689
% 1.75/1.92  clauses kept                 450
% 1.75/1.92  clauses forward subsumed     269
% 1.75/1.92  clauses back subsumed         26
% 1.75/1.92  Kbytes malloced             1953
% 1.75/1.92  
% 1.75/1.92  ----------- times (seconds) -----------
% 1.75/1.92  user CPU time          0.03          (0 hr, 0 min, 0 sec)
% 1.75/1.92  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.75/1.92  wall-clock time        2             (0 hr, 0 min, 2 sec)
% 1.75/1.92  
% 1.75/1.92  That finishes the proof of the theorem.
% 1.75/1.92  
% 1.75/1.92  Process 3662 finished Wed Jul 27 04:01:23 2022
% 1.75/1.92  Otter interrupted
% 1.75/1.92  PROOF FOUND
%------------------------------------------------------------------------------