TSTP Solution File: BOO075-1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : BOO075-1 : TPTP v8.1.0. Released v2.6.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n017.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 12:47:41 EDT 2022

% Result   : Unsatisfiable 1.65s 1.92s
% Output   : Refutation 1.65s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    7
%            Number of leaves      :    2
% Syntax   : Number of clauses     :   14 (  14 unt;   0 nHn;   2 RR)
%            Number of literals    :   14 (  13 equ;   1 neg)
%            Maximal clause size   :    1 (   1 avg)
%            Maximal term depth    :    8 (   2 avg)
%            Number of predicates  :    2 (   0 usr;   1 prp; 0-2 aty)
%            Number of functors    :    3 (   3 usr;   2 con; 0-2 aty)
%            Number of variables   :   29 (   4 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(1,axiom,
    nand(nand(a,a),nand(b,a)) != a,
    file('BOO075-1.p',unknown),
    [] ).

cnf(4,axiom,
    nand(nand(A,nand(nand(B,A),A)),nand(B,nand(C,A))) = B,
    file('BOO075-1.p',unknown),
    [] ).

cnf(5,plain,
    nand(nand(nand(A,nand(B,C)),nand(A,nand(A,nand(B,C)))),nand(nand(C,nand(nand(A,C),C)),nand(D,nand(A,nand(B,C))))) = nand(C,nand(nand(A,C),C)),
    inference(para_into,[status(thm),theory(equality)],[4,4]),
    [iquote('para_into,3.1.1.1.2.1,3.1.1')] ).

cnf(7,plain,
    nand(nand(nand(A,nand(B,C)),nand(nand(D,nand(A,nand(B,C))),nand(A,nand(B,C)))),nand(D,A)) = D,
    inference(para_into,[status(thm),theory(equality)],[4,4]),
    [iquote('para_into,3.1.1.2.2,3.1.1')] ).

cnf(9,plain,
    nand(nand(nand(A,B),nand(nand(nand(B,nand(nand(C,B),B)),nand(A,B)),nand(A,B))),C) = nand(B,nand(nand(C,B),B)),
    inference(para_into,[status(thm),theory(equality)],[4,4]),
    [iquote('para_into,3.1.1.2,3.1.1')] ).

cnf(16,plain,
    nand(nand(A,nand(nand(B,A),A)),nand(B,nand(C,nand(nand(A,C),C)))) = B,
    inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[7,4]),4,4]),
    [iquote('para_into,7.1.1.1.1,3.1.1,demod,4,4')] ).

cnf(49,plain,
    nand(nand(A,nand(nand(nand(B,nand(nand(B,B),B)),A),A)),B) = nand(B,nand(nand(B,B),B)),
    inference(para_into,[status(thm),theory(equality)],[16,4]),
    [iquote('para_into,15.1.1.2,3.1.1')] ).

cnf(80,plain,
    nand(A,nand(nand(A,A),A)) = nand(A,A),
    inference(flip,[status(thm),theory(equality)],[inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[5,16]),16])]),
    [iquote('para_into,5.1.1.2,15.1.1,demod,16,flip.1')] ).

cnf(83,plain,
    nand(nand(A,nand(nand(nand(B,B),A),A)),B) = nand(B,B),
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[49]),80,80]),
    [iquote('back_demod,49,demod,80,80')] ).

cnf(114,plain,
    nand(nand(A,A),nand(A,A)) = A,
    inference(demod,[status(thm),theory(equality)],[inference(para_from,[status(thm),theory(equality)],[80,4]),80]),
    [iquote('para_from,79.1.1,3.1.1.2,demod,80')] ).

cnf(119,plain,
    nand(nand(A,A),nand(A,nand(B,A))) = A,
    inference(para_from,[status(thm),theory(equality)],[80,4]),
    [iquote('para_from,79.1.1,3.1.1.1')] ).

cnf(245,plain,
    nand(A,nand(nand(nand(nand(B,A),nand(B,A)),A),A)) = nand(B,A),
    inference(flip,[status(thm),theory(equality)],[inference(demod,[status(thm),theory(equality)],[inference(para_from,[status(thm),theory(equality)],[83,9]),80,114])]),
    [iquote('para_from,83.1.1,9.1.1.1.2.1,demod,80,114,flip.1')] ).

cnf(443,plain,
    nand(nand(A,A),nand(B,A)) = A,
    inference(para_from,[status(thm),theory(equality)],[245,119]),
    [iquote('para_from,245.1.1,119.1.1.2')] ).

cnf(445,plain,
    $false,
    inference(binary,[status(thm)],[443,1]),
    [iquote('binary,443.1,1.1')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.11  % Problem  : BOO075-1 : TPTP v8.1.0. Released v2.6.0.
% 0.11/0.12  % Command  : otter-tptp-script %s
% 0.12/0.33  % Computer : n017.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 01:55:18 EDT 2022
% 0.12/0.33  % CPUTime  : 
% 1.65/1.92  ----- Otter 3.3f, August 2004 -----
% 1.65/1.92  The process was started by sandbox2 on n017.cluster.edu,
% 1.65/1.92  Wed Jul 27 01:55:18 2022
% 1.65/1.92  The command was "./otter".  The process ID is 1720.
% 1.65/1.92  
% 1.65/1.92  set(prolog_style_variables).
% 1.65/1.92  set(auto).
% 1.65/1.92     dependent: set(auto1).
% 1.65/1.92     dependent: set(process_input).
% 1.65/1.92     dependent: clear(print_kept).
% 1.65/1.92     dependent: clear(print_new_demod).
% 1.65/1.92     dependent: clear(print_back_demod).
% 1.65/1.92     dependent: clear(print_back_sub).
% 1.65/1.92     dependent: set(control_memory).
% 1.65/1.92     dependent: assign(max_mem, 12000).
% 1.65/1.92     dependent: assign(pick_given_ratio, 4).
% 1.65/1.92     dependent: assign(stats_level, 1).
% 1.65/1.92     dependent: assign(max_seconds, 10800).
% 1.65/1.92  clear(print_given).
% 1.65/1.92  
% 1.65/1.92  list(usable).
% 1.65/1.92  0 [] A=A.
% 1.65/1.92  0 [] nand(nand(A,nand(nand(B,A),A)),nand(B,nand(C,A)))=B.
% 1.65/1.92  0 [] nand(nand(a,a),nand(b,a))!=a.
% 1.65/1.92  end_of_list.
% 1.65/1.92  
% 1.65/1.92  SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.65/1.92  
% 1.65/1.92  All clauses are units, and equality is present; the
% 1.65/1.92  strategy will be Knuth-Bendix with positive clauses in sos.
% 1.65/1.92  
% 1.65/1.92     dependent: set(knuth_bendix).
% 1.65/1.92     dependent: set(anl_eq).
% 1.65/1.92     dependent: set(para_from).
% 1.65/1.92     dependent: set(para_into).
% 1.65/1.92     dependent: clear(para_from_right).
% 1.65/1.92     dependent: clear(para_into_right).
% 1.65/1.92     dependent: set(para_from_vars).
% 1.65/1.92     dependent: set(eq_units_both_ways).
% 1.65/1.92     dependent: set(dynamic_demod_all).
% 1.65/1.92     dependent: set(dynamic_demod).
% 1.65/1.92     dependent: set(order_eq).
% 1.65/1.92     dependent: set(back_demod).
% 1.65/1.92     dependent: set(lrpo).
% 1.65/1.92  
% 1.65/1.92  ------------> process usable:
% 1.65/1.92  ** KEPT (pick-wt=9): 1 [] nand(nand(a,a),nand(b,a))!=a.
% 1.65/1.92  
% 1.65/1.92  ------------> process sos:
% 1.65/1.92  ** KEPT (pick-wt=3): 2 [] A=A.
% 1.65/1.92  ** KEPT (pick-wt=15): 3 [] nand(nand(A,nand(nand(B,A),A)),nand(B,nand(C,A)))=B.
% 1.65/1.92  ---> New Demodulator: 4 [new_demod,3] nand(nand(A,nand(nand(B,A),A)),nand(B,nand(C,A)))=B.
% 1.65/1.92    Following clause subsumed by 2 during input processing: 0 [copy,2,flip.1] A=A.
% 1.65/1.92  >>>> Starting back demodulation with 4.
% 1.65/1.92  
% 1.65/1.92  ======= end of input processing =======
% 1.65/1.92  
% 1.65/1.92  =========== start of search ===========
% 1.65/1.92  
% 1.65/1.92  -------- PROOF -------- 
% 1.65/1.92  
% 1.65/1.92  ----> UNIT CONFLICT at   0.06 sec ----> 445 [binary,443.1,1.1] $F.
% 1.65/1.92  
% 1.65/1.92  Length of proof is 11.  Level of proof is 6.
% 1.65/1.92  
% 1.65/1.92  ---------------- PROOF ----------------
% 1.65/1.92  % SZS status Unsatisfiable
% 1.65/1.92  % SZS output start Refutation
% See solution above
% 1.65/1.92  ------------ end of proof -------------
% 1.65/1.92  
% 1.65/1.92  
% 1.65/1.92  Search stopped by max_proofs option.
% 1.65/1.92  
% 1.65/1.92  
% 1.65/1.92  Search stopped by max_proofs option.
% 1.65/1.92  
% 1.65/1.92  ============ end of search ============
% 1.65/1.92  
% 1.65/1.92  -------------- statistics -------------
% 1.65/1.92  clauses given                 17
% 1.65/1.92  clauses generated            668
% 1.65/1.92  clauses kept                 223
% 1.65/1.92  clauses forward subsumed     536
% 1.65/1.92  clauses back subsumed          0
% 1.65/1.92  Kbytes malloced             2929
% 1.65/1.92  
% 1.65/1.92  ----------- times (seconds) -----------
% 1.65/1.92  user CPU time          0.06          (0 hr, 0 min, 0 sec)
% 1.65/1.92  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.65/1.92  wall-clock time        1             (0 hr, 0 min, 1 sec)
% 1.65/1.92  
% 1.65/1.92  That finishes the proof of the theorem.
% 1.65/1.92  
% 1.65/1.92  Process 1720 finished Wed Jul 27 01:55:19 2022
% 1.74/1.92  Otter interrupted
% 1.74/1.92  PROOF FOUND
%------------------------------------------------------------------------------