TSTP Solution File: SYN083-1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : SYN083-1 : TPTP v8.1.0. Released v1.0.0.
% 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 13:23:32 EDT 2022

% Result   : Unsatisfiable 1.65s 1.84s
% Output   : Refutation 1.65s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    3
%            Number of leaves      :    3
% Syntax   : Number of clauses     :    7 (   7 unt;   0 nHn;   4 RR)
%            Number of literals    :    7 (   6 equ;   3 neg)
%            Maximal clause size   :    1 (   1 avg)
%            Maximal term depth    :    4 (   2 avg)
%            Number of predicates  :    2 (   0 usr;   1 prp; 0-2 aty)
%            Number of functors    :    5 (   5 usr;   4 con; 0-2 aty)
%            Number of variables   :    7 (   0 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(1,axiom,
    f(a,f(b,f(c,d))) != f(f(f(a,b),c),d),
    file('SYN083-1.p',unknown),
    [] ).

cnf(2,plain,
    f(f(f(a,b),c),d) != f(a,f(b,f(c,d))),
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[1])]),
    [iquote('copy,1,flip.1')] ).

cnf(3,axiom,
    A = A,
    file('SYN083-1.p',unknown),
    [] ).

cnf(4,axiom,
    f(A,f(B,C)) = f(f(A,B),C),
    file('SYN083-1.p',unknown),
    [] ).

cnf(6,plain,
    f(f(A,B),C) = f(A,f(B,C)),
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[4])]),
    [iquote('copy,4,flip.1')] ).

cnf(7,plain,
    f(a,f(b,f(c,d))) != f(a,f(b,f(c,d))),
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[2]),6,6,6]),
    [iquote('back_demod,2,demod,6,6,6')] ).

cnf(8,plain,
    $false,
    inference(binary,[status(thm)],[7,3]),
    [iquote('binary,7.1,3.1')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.11  % Problem  : SYN083-1 : TPTP v8.1.0. Released v1.0.0.
% 0.07/0.12  % 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 11:00:35 EDT 2022
% 0.12/0.33  % CPUTime  : 
% 1.65/1.84  
% 1.65/1.84  -------- PROOF -------- 
% 1.65/1.84  ----- Otter 3.3f, August 2004 -----
% 1.65/1.84  The process was started by sandbox on n012.cluster.edu,
% 1.65/1.84  Wed Jul 27 11:00:35 2022
% 1.65/1.84  The command was "./otter".  The process ID is 26852.
% 1.65/1.84  
% 1.65/1.84  set(prolog_style_variables).
% 1.65/1.84  set(auto).
% 1.65/1.84     dependent: set(auto1).
% 1.65/1.84     dependent: set(process_input).
% 1.65/1.84     dependent: clear(print_kept).
% 1.65/1.84     dependent: clear(print_new_demod).
% 1.65/1.84     dependent: clear(print_back_demod).
% 1.65/1.84     dependent: clear(print_back_sub).
% 1.65/1.84     dependent: set(control_memory).
% 1.65/1.84     dependent: assign(max_mem, 12000).
% 1.65/1.84     dependent: assign(pick_given_ratio, 4).
% 1.65/1.84     dependent: assign(stats_level, 1).
% 1.65/1.84     dependent: assign(max_seconds, 10800).
% 1.65/1.84  clear(print_given).
% 1.65/1.84  
% 1.65/1.84  list(usable).
% 1.65/1.84  0 [] A=A.
% 1.65/1.84  0 [] f(X,f(Y,Z))=f(f(X,Y),Z).
% 1.65/1.84  0 [] f(a,f(b,f(c,d)))!=f(f(f(a,b),c),d).
% 1.65/1.84  end_of_list.
% 1.65/1.84  
% 1.65/1.84  SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.65/1.84  
% 1.65/1.84  All clauses are units, and equality is present; the
% 1.65/1.84  strategy will be Knuth-Bendix with positive clauses in sos.
% 1.65/1.84  
% 1.65/1.84     dependent: set(knuth_bendix).
% 1.65/1.84     dependent: set(anl_eq).
% 1.65/1.84     dependent: set(para_from).
% 1.65/1.84     dependent: set(para_into).
% 1.65/1.84     dependent: clear(para_from_right).
% 1.65/1.84     dependent: clear(para_into_right).
% 1.65/1.84     dependent: set(para_from_vars).
% 1.65/1.84     dependent: set(eq_units_both_ways).
% 1.65/1.84     dependent: set(dynamic_demod_all).
% 1.65/1.84     dependent: set(dynamic_demod).
% 1.65/1.84     dependent: set(order_eq).
% 1.65/1.84     dependent: set(back_demod).
% 1.65/1.84     dependent: set(lrpo).
% 1.65/1.84  
% 1.65/1.84  ------------> process usable:
% 1.65/1.84  ** KEPT (pick-wt=15): 2 [copy,1,flip.1] f(f(f(a,b),c),d)!=f(a,f(b,f(c,d))).
% 1.65/1.84  
% 1.65/1.84  ------------> process sos:
% 1.65/1.84  ** KEPT (pick-wt=3): 3 [] A=A.
% 1.65/1.84  ** KEPT (pick-wt=11): 5 [copy,4,flip.1] f(f(A,B),C)=f(A,f(B,C)).
% 1.65/1.84  ---> New Demodulator: 6 [new_demod,5] f(f(A,B),C)=f(A,f(B,C)).
% 1.65/1.84    Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.65/1.84  >>>> Starting back demodulation with 6.
% 1.65/1.84      >> back demodulating 2 with 6.
% 1.65/1.84  
% 1.65/1.84  ----> UNIT CONFLICT at   0.00 sec ----> 8 [binary,7.1,3.1] $F.
% 1.65/1.84  
% 1.65/1.84  Length of proof is 3.  Level of proof is 2.
% 1.65/1.84  
% 1.65/1.84  ---------------- PROOF ----------------
% 1.65/1.84  % SZS status Unsatisfiable
% 1.65/1.84  % SZS output start Refutation
% See solution above
% 1.65/1.84  ------------ end of proof -------------
% 1.65/1.84  
% 1.65/1.84  
% 1.65/1.84  Search stopped by max_proofs option.
% 1.65/1.84  
% 1.65/1.84  
% 1.65/1.84  Search stopped by max_proofs option.
% 1.65/1.84  
% 1.65/1.84  ============ end of search ============
% 1.65/1.84  
% 1.65/1.84  -------------- statistics -------------
% 1.65/1.84  clauses given                  0
% 1.65/1.84  clauses generated              0
% 1.65/1.84  clauses kept                   4
% 1.65/1.84  clauses forward subsumed       1
% 1.65/1.84  clauses back subsumed          0
% 1.65/1.84  Kbytes malloced              976
% 1.65/1.84  
% 1.65/1.84  ----------- times (seconds) -----------
% 1.65/1.84  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.65/1.84  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.65/1.84  wall-clock time        2             (0 hr, 0 min, 2 sec)
% 1.65/1.84  
% 1.65/1.84  That finishes the proof of the theorem.
% 1.65/1.84  
% 1.65/1.84  Process 26852 finished Wed Jul 27 11:00:37 2022
% 1.65/1.84  Otter interrupted
% 1.65/1.84  PROOF FOUND
%------------------------------------------------------------------------------