TSTP Solution File: GRP549-1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : GRP549-1 : TPTP v8.1.0. Released v2.6.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n014.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 12:57:14 EDT 2022

% Result   : Unsatisfiable 1.82s 2.00s
% Output   : Refutation 1.82s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    3
%            Number of leaves      :    5
% Syntax   : Number of clauses     :    9 (   9 unt;   0 nHn;   4 RR)
%            Number of literals    :    9 (   8 equ;   3 neg)
%            Maximal clause size   :    1 (   1 avg)
%            Maximal term depth    :    3 (   1 avg)
%            Number of predicates  :    2 (   0 usr;   1 prp; 0-2 aty)
%            Number of functors    :    6 (   6 usr;   3 con; 0-2 aty)
%            Number of variables   :    6 (   0 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(1,axiom,
    multiply(inverse(a1),a1) != multiply(inverse(b1),b1),
    file('GRP549-1.p',unknown),
    [] ).

cnf(2,plain,
    multiply(inverse(b1),b1) != multiply(inverse(a1),a1),
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[1])]),
    [iquote('copy,1,flip.1')] ).

cnf(3,axiom,
    A = A,
    file('GRP549-1.p',unknown),
    [] ).

cnf(7,axiom,
    multiply(A,B) = divide(A,divide(identity,B)),
    file('GRP549-1.p',unknown),
    [] ).

cnf(9,axiom,
    inverse(A) = divide(identity,A),
    file('GRP549-1.p',unknown),
    [] ).

cnf(10,axiom,
    identity = divide(A,A),
    file('GRP549-1.p',unknown),
    [] ).

cnf(12,plain,
    divide(A,A) = identity,
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[10])]),
    [iquote('copy,10,flip.1')] ).

cnf(13,plain,
    identity != identity,
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[2]),9,7,12,9,7,12]),
    [iquote('back_demod,2,demod,9,7,12,9,7,12')] ).

cnf(14,plain,
    $false,
    inference(binary,[status(thm)],[13,3]),
    [iquote('binary,13.1,3.1')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem  : GRP549-1 : TPTP v8.1.0. Released v2.6.0.
% 0.07/0.13  % Command  : otter-tptp-script %s
% 0.13/0.34  % Computer : n014.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 05:31:24 EDT 2022
% 0.13/0.35  % CPUTime  : 
% 1.82/2.00  
% 1.82/2.00  -------- PROOF -------- 
% 1.82/2.00  ----- Otter 3.3f, August 2004 -----
% 1.82/2.00  The process was started by sandbox on n014.cluster.edu,
% 1.82/2.00  Wed Jul 27 05:31:24 2022
% 1.82/2.00  The command was "./otter".  The process ID is 14379.
% 1.82/2.00  
% 1.82/2.00  set(prolog_style_variables).
% 1.82/2.00  set(auto).
% 1.82/2.00     dependent: set(auto1).
% 1.82/2.00     dependent: set(process_input).
% 1.82/2.00     dependent: clear(print_kept).
% 1.82/2.00     dependent: clear(print_new_demod).
% 1.82/2.00     dependent: clear(print_back_demod).
% 1.82/2.00     dependent: clear(print_back_sub).
% 1.82/2.00     dependent: set(control_memory).
% 1.82/2.00     dependent: assign(max_mem, 12000).
% 1.82/2.00     dependent: assign(pick_given_ratio, 4).
% 1.82/2.00     dependent: assign(stats_level, 1).
% 1.82/2.00     dependent: assign(max_seconds, 10800).
% 1.82/2.00  clear(print_given).
% 1.82/2.00  
% 1.82/2.00  list(usable).
% 1.82/2.00  0 [] A=A.
% 1.82/2.00  0 [] divide(divide(identity,A),divide(divide(divide(B,A),C),B))=C.
% 1.82/2.00  0 [] multiply(A,B)=divide(A,divide(identity,B)).
% 1.82/2.00  0 [] inverse(A)=divide(identity,A).
% 1.82/2.00  0 [] identity=divide(A,A).
% 1.82/2.00  0 [] multiply(inverse(a1),a1)!=multiply(inverse(b1),b1).
% 1.82/2.00  end_of_list.
% 1.82/2.00  
% 1.82/2.00  SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.82/2.00  
% 1.82/2.00  All clauses are units, and equality is present; the
% 1.82/2.00  strategy will be Knuth-Bendix with positive clauses in sos.
% 1.82/2.00  
% 1.82/2.00     dependent: set(knuth_bendix).
% 1.82/2.00     dependent: set(anl_eq).
% 1.82/2.00     dependent: set(para_from).
% 1.82/2.00     dependent: set(para_into).
% 1.82/2.00     dependent: clear(para_from_right).
% 1.82/2.00     dependent: clear(para_into_right).
% 1.82/2.00     dependent: set(para_from_vars).
% 1.82/2.00     dependent: set(eq_units_both_ways).
% 1.82/2.00     dependent: set(dynamic_demod_all).
% 1.82/2.00     dependent: set(dynamic_demod).
% 1.82/2.00     dependent: set(order_eq).
% 1.82/2.00     dependent: set(back_demod).
% 1.82/2.00     dependent: set(lrpo).
% 1.82/2.00  
% 1.82/2.00  ------------> process usable:
% 1.82/2.00  ** KEPT (pick-wt=9): 2 [copy,1,flip.1] multiply(inverse(b1),b1)!=multiply(inverse(a1),a1).
% 1.82/2.00  
% 1.82/2.00  ------------> process sos:
% 1.82/2.00  ** KEPT (pick-wt=3): 3 [] A=A.
% 1.82/2.00  ** KEPT (pick-wt=13): 4 [] divide(divide(identity,A),divide(divide(divide(B,A),C),B))=C.
% 1.82/2.00  ---> New Demodulator: 5 [new_demod,4] divide(divide(identity,A),divide(divide(divide(B,A),C),B))=C.
% 1.82/2.00  ** KEPT (pick-wt=9): 6 [] multiply(A,B)=divide(A,divide(identity,B)).
% 1.82/2.00  ---> New Demodulator: 7 [new_demod,6] multiply(A,B)=divide(A,divide(identity,B)).
% 1.82/2.00  ** KEPT (pick-wt=6): 8 [] inverse(A)=divide(identity,A).
% 1.82/2.00  ---> New Demodulator: 9 [new_demod,8] inverse(A)=divide(identity,A).
% 1.82/2.00  ** KEPT (pick-wt=5): 11 [copy,10,flip.1] divide(A,A)=identity.
% 1.82/2.00  ---> New Demodulator: 12 [new_demod,11] divide(A,A)=identity.
% 1.82/2.00    Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.82/2.00  >>>> Starting back demodulation with 5.
% 1.82/2.00  >>>> Starting back demodulation with 7.
% 1.82/2.00      >> back demodulating 2 with 7.
% 1.82/2.00  
% 1.82/2.00  ----> UNIT CONFLICT at   0.00 sec ----> 14 [binary,13.1,3.1] $F.
% 1.82/2.00  
% 1.82/2.00  Length of proof is 3.  Level of proof is 2.
% 1.82/2.00  
% 1.82/2.00  ---------------- PROOF ----------------
% 1.82/2.00  % SZS status Unsatisfiable
% 1.82/2.00  % SZS output start Refutation
% See solution above
% 1.82/2.00  ------------ end of proof -------------
% 1.82/2.00  
% 1.82/2.00  
% 1.82/2.00  Search stopped by max_proofs option.
% 1.82/2.00  
% 1.82/2.00  
% 1.82/2.00  Search stopped by max_proofs option.
% 1.82/2.00  
% 1.82/2.00  ============ end of search ============
% 1.82/2.00  
% 1.82/2.00  -------------- statistics -------------
% 1.82/2.00  clauses given                  0
% 1.82/2.00  clauses generated              0
% 1.82/2.00  clauses kept                   7
% 1.82/2.00  clauses forward subsumed       1
% 1.82/2.00  clauses back subsumed          0
% 1.82/2.00  Kbytes malloced              976
% 1.82/2.00  
% 1.82/2.00  ----------- times (seconds) -----------
% 1.82/2.00  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.82/2.00  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.82/2.00  wall-clock time        2             (0 hr, 0 min, 2 sec)
% 1.82/2.00  
% 1.82/2.00  That finishes the proof of the theorem.
% 1.82/2.00  
% 1.82/2.00  Process 14379 finished Wed Jul 27 05:31:26 2022
% 1.82/2.00  Otter interrupted
% 1.82/2.00  PROOF FOUND
%------------------------------------------------------------------------------