TSTP Solution File: GRP457-1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : GRP457-1 : TPTP v8.1.0. Released v2.6.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n015.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 12:57:03 EDT 2022

% Result   : Unsatisfiable 1.80s 2.02s
% Output   : Refutation 1.80s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    3
%            Number of leaves      :    5
% Syntax   : Number of clauses     :    8 (   8 unt;   0 nHn;   3 RR)
%            Number of literals    :    8 (   7 equ;   2 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    :    5 (   5 usr;   2 con; 0-2 aty)
%            Number of variables   :    6 (   0 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(1,axiom,
    multiply(inverse(a1),a1) != identity,
    file('GRP457-1.p',unknown),
    [] ).

cnf(2,axiom,
    A = A,
    file('GRP457-1.p',unknown),
    [] ).

cnf(6,axiom,
    multiply(A,B) = divide(A,divide(identity,B)),
    file('GRP457-1.p',unknown),
    [] ).

cnf(8,axiom,
    inverse(A) = divide(identity,A),
    file('GRP457-1.p',unknown),
    [] ).

cnf(9,axiom,
    identity = divide(A,A),
    file('GRP457-1.p',unknown),
    [] ).

cnf(11,plain,
    divide(A,A) = identity,
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[9])]),
    [iquote('copy,9,flip.1')] ).

cnf(12,plain,
    identity != identity,
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[1]),8,6,11]),
    [iquote('back_demod,1,demod,8,6,11')] ).

cnf(13,plain,
    $false,
    inference(binary,[status(thm)],[12,2]),
    [iquote('binary,12.1,2.1')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12  % Problem  : GRP457-1 : TPTP v8.1.0. Released v2.6.0.
% 0.11/0.12  % Command  : otter-tptp-script %s
% 0.12/0.33  % Computer : n015.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 05:10:11 EDT 2022
% 0.12/0.33  % CPUTime  : 
% 1.80/2.02  
% 1.80/2.02  -------- PROOF -------- 
% 1.80/2.02  ----- Otter 3.3f, August 2004 -----
% 1.80/2.02  The process was started by sandbox2 on n015.cluster.edu,
% 1.80/2.02  Wed Jul 27 05:10:11 2022
% 1.80/2.02  The command was "./otter".  The process ID is 11670.
% 1.80/2.02  
% 1.80/2.02  set(prolog_style_variables).
% 1.80/2.02  set(auto).
% 1.80/2.02     dependent: set(auto1).
% 1.80/2.02     dependent: set(process_input).
% 1.80/2.02     dependent: clear(print_kept).
% 1.80/2.02     dependent: clear(print_new_demod).
% 1.80/2.02     dependent: clear(print_back_demod).
% 1.80/2.02     dependent: clear(print_back_sub).
% 1.80/2.02     dependent: set(control_memory).
% 1.80/2.02     dependent: assign(max_mem, 12000).
% 1.80/2.02     dependent: assign(pick_given_ratio, 4).
% 1.80/2.02     dependent: assign(stats_level, 1).
% 1.80/2.02     dependent: assign(max_seconds, 10800).
% 1.80/2.02  clear(print_given).
% 1.80/2.02  
% 1.80/2.02  list(usable).
% 1.80/2.02  0 [] A=A.
% 1.80/2.02  0 [] divide(divide(divide(A,A),divide(A,divide(B,divide(divide(identity,A),C)))),C)=B.
% 1.80/2.02  0 [] multiply(A,B)=divide(A,divide(identity,B)).
% 1.80/2.02  0 [] inverse(A)=divide(identity,A).
% 1.80/2.02  0 [] identity=divide(A,A).
% 1.80/2.02  0 [] multiply(inverse(a1),a1)!=identity.
% 1.80/2.02  end_of_list.
% 1.80/2.02  
% 1.80/2.02  SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.80/2.02  
% 1.80/2.02  All clauses are units, and equality is present; the
% 1.80/2.02  strategy will be Knuth-Bendix with positive clauses in sos.
% 1.80/2.02  
% 1.80/2.02     dependent: set(knuth_bendix).
% 1.80/2.02     dependent: set(anl_eq).
% 1.80/2.02     dependent: set(para_from).
% 1.80/2.02     dependent: set(para_into).
% 1.80/2.02     dependent: clear(para_from_right).
% 1.80/2.02     dependent: clear(para_into_right).
% 1.80/2.02     dependent: set(para_from_vars).
% 1.80/2.02     dependent: set(eq_units_both_ways).
% 1.80/2.02     dependent: set(dynamic_demod_all).
% 1.80/2.02     dependent: set(dynamic_demod).
% 1.80/2.02     dependent: set(order_eq).
% 1.80/2.02     dependent: set(back_demod).
% 1.80/2.02     dependent: set(lrpo).
% 1.80/2.02  
% 1.80/2.02  ------------> process usable:
% 1.80/2.02  ** KEPT (pick-wt=6): 1 [] multiply(inverse(a1),a1)!=identity.
% 1.80/2.02  
% 1.80/2.02  ------------> process sos:
% 1.80/2.02  ** KEPT (pick-wt=3): 2 [] A=A.
% 1.80/2.02  ** KEPT (pick-wt=17): 3 [] divide(divide(divide(A,A),divide(A,divide(B,divide(divide(identity,A),C)))),C)=B.
% 1.80/2.02  ---> New Demodulator: 4 [new_demod,3] divide(divide(divide(A,A),divide(A,divide(B,divide(divide(identity,A),C)))),C)=B.
% 1.80/2.02  ** KEPT (pick-wt=9): 5 [] multiply(A,B)=divide(A,divide(identity,B)).
% 1.80/2.02  ---> New Demodulator: 6 [new_demod,5] multiply(A,B)=divide(A,divide(identity,B)).
% 1.80/2.02  ** KEPT (pick-wt=6): 7 [] inverse(A)=divide(identity,A).
% 1.80/2.02  ---> New Demodulator: 8 [new_demod,7] inverse(A)=divide(identity,A).
% 1.80/2.02  ** KEPT (pick-wt=5): 10 [copy,9,flip.1] divide(A,A)=identity.
% 1.80/2.02  ---> New Demodulator: 11 [new_demod,10] divide(A,A)=identity.
% 1.80/2.02    Following clause subsumed by 2 during input processing: 0 [copy,2,flip.1] A=A.
% 1.80/2.02  >>>> Starting back demodulation with 4.
% 1.80/2.02  >>>> Starting back demodulation with 6.
% 1.80/2.02      >> back demodulating 1 with 6.
% 1.80/2.02  
% 1.80/2.02  ----> UNIT CONFLICT at   0.00 sec ----> 13 [binary,12.1,2.1] $F.
% 1.80/2.02  
% 1.80/2.02  Length of proof is 2.  Level of proof is 2.
% 1.80/2.02  
% 1.80/2.02  ---------------- PROOF ----------------
% 1.80/2.02  % SZS status Unsatisfiable
% 1.80/2.02  % SZS output start Refutation
% See solution above
% 1.80/2.02  ------------ end of proof -------------
% 1.80/2.02  
% 1.80/2.02  
% 1.80/2.02  Search stopped by max_proofs option.
% 1.80/2.02  
% 1.80/2.02  
% 1.80/2.02  Search stopped by max_proofs option.
% 1.80/2.02  
% 1.80/2.02  ============ end of search ============
% 1.80/2.02  
% 1.80/2.02  -------------- statistics -------------
% 1.80/2.02  clauses given                  0
% 1.80/2.02  clauses generated              0
% 1.80/2.02  clauses kept                   7
% 1.80/2.02  clauses forward subsumed       1
% 1.80/2.02  clauses back subsumed          0
% 1.80/2.02  Kbytes malloced              976
% 1.80/2.02  
% 1.80/2.02  ----------- times (seconds) -----------
% 1.80/2.02  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.80/2.02  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.80/2.02  wall-clock time        2             (0 hr, 0 min, 2 sec)
% 1.80/2.02  
% 1.80/2.02  That finishes the proof of the theorem.
% 1.80/2.02  
% 1.80/2.02  Process 11670 finished Wed Jul 27 05:10:13 2022
% 1.80/2.02  Otter interrupted
% 1.80/2.02  PROOF FOUND
%------------------------------------------------------------------------------