TSTP Solution File: SYN950+1 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : SYN950+1 : TPTP v8.1.0. Released v3.1.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n026.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 13:25:33 EDT 2022

% Result   : Theorem 1.89s 2.05s
% Output   : Refutation 1.89s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    2
%            Number of leaves      :    3
% Syntax   : Number of clauses     :    5 (   2 unt;   1 nHn;   3 RR)
%            Number of literals    :    8 (   0 equ;   3 neg)
%            Maximal clause size   :    2 (   1 avg)
%            Maximal term depth    :    1 (   1 avg)
%            Number of predicates  :    3 (   2 usr;   1 prp; 0-1 aty)
%            Number of functors    :    2 (   2 usr;   2 con; 0-0 aty)
%            Number of variables   :    3 (   1 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(1,axiom,
    ( ~ q(A)
    | p(A) ),
    file('SYN950+1.p',unknown),
    [] ).

cnf(4,axiom,
    ( ~ p(dollar_c2)
    | ~ p(dollar_c1) ),
    file('SYN950+1.p',unknown),
    [] ).

cnf(5,axiom,
    ( p(A)
    | q(A) ),
    file('SYN950+1.p',unknown),
    [] ).

cnf(6,plain,
    p(A),
    inference(factor_simp,[status(thm)],[inference(hyper,[status(thm)],[5,1])]),
    [iquote('hyper,5,1,factor_simp')] ).

cnf(7,plain,
    $false,
    inference(hyper,[status(thm)],[6,4,6]),
    [iquote('hyper,6,4,6')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem  : SYN950+1 : TPTP v8.1.0. Released v3.1.0.
% 0.07/0.13  % Command  : otter-tptp-script %s
% 0.13/0.34  % Computer : n026.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 11:38:09 EDT 2022
% 0.13/0.34  % CPUTime  : 
% 1.89/2.05  ----- Otter 3.3f, August 2004 -----
% 1.89/2.05  The process was started by sandbox2 on n026.cluster.edu,
% 1.89/2.05  Wed Jul 27 11:38:09 2022
% 1.89/2.05  The command was "./otter".  The process ID is 1973.
% 1.89/2.05  
% 1.89/2.05  set(prolog_style_variables).
% 1.89/2.05  set(auto).
% 1.89/2.05     dependent: set(auto1).
% 1.89/2.05     dependent: set(process_input).
% 1.89/2.05     dependent: clear(print_kept).
% 1.89/2.05     dependent: clear(print_new_demod).
% 1.89/2.05     dependent: clear(print_back_demod).
% 1.89/2.05     dependent: clear(print_back_sub).
% 1.89/2.05     dependent: set(control_memory).
% 1.89/2.05     dependent: assign(max_mem, 12000).
% 1.89/2.05     dependent: assign(pick_given_ratio, 4).
% 1.89/2.05     dependent: assign(stats_level, 1).
% 1.89/2.05     dependent: assign(max_seconds, 10800).
% 1.89/2.05  clear(print_given).
% 1.89/2.05  
% 1.89/2.05  formula_list(usable).
% 1.89/2.05  -(all A B ((all Y (q(Y)->p(Y)))-> (exists X ((p(X)->p(A))& (q(X)->p(B)))))).
% 1.89/2.05  end_of_list.
% 1.89/2.05  
% 1.89/2.05  -------> usable clausifies to:
% 1.89/2.05  
% 1.89/2.05  list(usable).
% 1.89/2.05  0 [] -q(Y)|p(Y).
% 1.89/2.05  0 [] p(X)|q(X).
% 1.89/2.05  0 [] p(X)| -p($c1).
% 1.89/2.05  0 [] -p($c2)|q(X).
% 1.89/2.05  0 [] -p($c2)| -p($c1).
% 1.89/2.05  end_of_list.
% 1.89/2.05  
% 1.89/2.05  SCAN INPUT: prop=0, horn=0, equality=0, symmetry=0, max_lits=2.
% 1.89/2.05  
% 1.89/2.05  This is a non-Horn set without equality.  The strategy will
% 1.89/2.05  be ordered hyper_res, unit deletion, and factoring, with
% 1.89/2.05  satellites in sos and with nuclei in usable.
% 1.89/2.05  
% 1.89/2.05     dependent: set(hyper_res).
% 1.89/2.05     dependent: set(factor).
% 1.89/2.05     dependent: set(unit_deletion).
% 1.89/2.05  
% 1.89/2.05  ------------> process usable:
% 1.89/2.05  ** KEPT (pick-wt=4): 1 [] -q(A)|p(A).
% 1.89/2.05  ** KEPT (pick-wt=4): 2 [] p(A)| -p($c1).
% 1.89/2.05  ** KEPT (pick-wt=4): 3 [] -p($c2)|q(A).
% 1.89/2.05  ** KEPT (pick-wt=4): 4 [] -p($c2)| -p($c1).
% 1.89/2.05  
% 1.89/2.05  ------------> process sos:
% 1.89/2.05  ** KEPT (pick-wt=4): 5 [] p(A)|q(A).
% 1.89/2.05  
% 1.89/2.05  ======= end of input processing =======
% 1.89/2.05  
% 1.89/2.05  =========== start of search ===========
% 1.89/2.05  
% 1.89/2.05  -------- PROOF -------- 
% 1.89/2.05  
% 1.89/2.05  -----> EMPTY CLAUSE at   0.00 sec ----> 7 [hyper,6,4,6] $F.
% 1.89/2.05  
% 1.89/2.05  Length of proof is 1.  Level of proof is 1.
% 1.89/2.05  
% 1.89/2.05  ---------------- PROOF ----------------
% 1.89/2.05  % SZS status Theorem
% 1.89/2.05  % SZS output start Refutation
% See solution above
% 1.89/2.05  ------------ end of proof -------------
% 1.89/2.05  
% 1.89/2.05  
% 1.89/2.05  Search stopped by max_proofs option.
% 1.89/2.05  
% 1.89/2.05  
% 1.89/2.05  Search stopped by max_proofs option.
% 1.89/2.05  
% 1.89/2.05  ============ end of search ============
% 1.89/2.05  
% 1.89/2.05  -------------- statistics -------------
% 1.89/2.05  clauses given                  2
% 1.89/2.05  clauses generated              2
% 1.89/2.05  clauses kept                   6
% 1.89/2.05  clauses forward subsumed       0
% 1.89/2.05  clauses back subsumed          3
% 1.89/2.05  Kbytes malloced              976
% 1.89/2.05  
% 1.89/2.05  ----------- times (seconds) -----------
% 1.89/2.05  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.89/2.05  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.89/2.05  wall-clock time        2             (0 hr, 0 min, 2 sec)
% 1.89/2.05  
% 1.89/2.05  That finishes the proof of the theorem.
% 1.89/2.05  
% 1.89/2.05  Process 1973 finished Wed Jul 27 11:38:11 2022
% 1.89/2.05  Otter interrupted
% 1.89/2.05  PROOF FOUND
%------------------------------------------------------------------------------