TSTP Solution File: LCL686+1.001 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : LCL686+1.001 : TPTP v8.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n011.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 13:05:10 EDT 2022

% Result   : Theorem 1.98s 2.16s
% Output   : Refutation 1.98s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    4
%            Number of leaves      :    6
% Syntax   : Number of clauses     :   11 (   5 unt;   2 nHn;  10 RR)
%            Number of literals    :   23 (   0 equ;  11 neg)
%            Maximal clause size   :    4 (   2 avg)
%            Maximal term depth    :    2 (   1 avg)
%            Number of predicates  :    4 (   3 usr;   1 prp; 0-2 aty)
%            Number of functors    :    2 (   2 usr;   1 con; 0-1 aty)
%            Number of variables   :    8 (   0 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(3,axiom,
    ( ~ r1(dollar_c3,A)
    | ~ r1(A,B)
    | ~ p2(B)
    | p1(B) ),
    file('LCL686+1.001.p',unknown),
    [] ).

cnf(4,axiom,
    ( ~ r1(dollar_c3,A)
    | ~ r1(A,B)
    | p2(B)
    | ~ p1(B) ),
    file('LCL686+1.001.p',unknown),
    [] ).

cnf(7,axiom,
    ( ~ r1(dollar_c3,A)
    | r1(A,dollar_f3(A)) ),
    file('LCL686+1.001.p',unknown),
    [] ).

cnf(8,axiom,
    ( ~ r1(dollar_c3,A)
    | p1(dollar_f3(A))
    | p2(dollar_f3(A)) ),
    file('LCL686+1.001.p',unknown),
    [] ).

cnf(9,axiom,
    ( ~ r1(dollar_c3,A)
    | ~ p2(dollar_f3(A))
    | ~ p1(dollar_f3(A)) ),
    file('LCL686+1.001.p',unknown),
    [] ).

cnf(12,axiom,
    r1(A,A),
    file('LCL686+1.001.p',unknown),
    [] ).

cnf(18,plain,
    ( p1(dollar_f3(dollar_c3))
    | p2(dollar_f3(dollar_c3)) ),
    inference(hyper,[status(thm)],[12,8]),
    [iquote('hyper,12,8')] ).

cnf(19,plain,
    r1(dollar_c3,dollar_f3(dollar_c3)),
    inference(hyper,[status(thm)],[12,7]),
    [iquote('hyper,12,7')] ).

cnf(38,plain,
    p1(dollar_f3(dollar_c3)),
    inference(factor_simp,[status(thm)],[inference(hyper,[status(thm)],[18,3,19,12])]),
    [iquote('hyper,18,3,19,12,factor_simp')] ).

cnf(39,plain,
    p2(dollar_f3(dollar_c3)),
    inference(hyper,[status(thm)],[38,4,19,12]),
    [iquote('hyper,38,4,19,12')] ).

cnf(40,plain,
    $false,
    inference(hyper,[status(thm)],[39,9,12,38]),
    [iquote('hyper,39,9,12,38')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12  % Problem  : LCL686+1.001 : TPTP v8.1.0. Released v4.0.0.
% 0.03/0.13  % Command  : otter-tptp-script %s
% 0.12/0.34  % Computer : n011.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit : 300
% 0.12/0.34  % WCLimit  : 300
% 0.12/0.34  % DateTime : Wed Jul 27 09:29:10 EDT 2022
% 0.12/0.34  % CPUTime  : 
% 1.98/2.16  ----- Otter 3.3f, August 2004 -----
% 1.98/2.16  The process was started by sandbox2 on n011.cluster.edu,
% 1.98/2.16  Wed Jul 27 09:29:10 2022
% 1.98/2.16  The command was "./otter".  The process ID is 18840.
% 1.98/2.16  
% 1.98/2.16  set(prolog_style_variables).
% 1.98/2.16  set(auto).
% 1.98/2.16     dependent: set(auto1).
% 1.98/2.16     dependent: set(process_input).
% 1.98/2.16     dependent: clear(print_kept).
% 1.98/2.16     dependent: clear(print_new_demod).
% 1.98/2.16     dependent: clear(print_back_demod).
% 1.98/2.16     dependent: clear(print_back_sub).
% 1.98/2.16     dependent: set(control_memory).
% 1.98/2.16     dependent: assign(max_mem, 12000).
% 1.98/2.16     dependent: assign(pick_given_ratio, 4).
% 1.98/2.16     dependent: assign(stats_level, 1).
% 1.98/2.16     dependent: assign(max_seconds, 10800).
% 1.98/2.16  clear(print_given).
% 1.98/2.16  
% 1.98/2.16  formula_list(usable).
% 1.98/2.16  all X r1(X,X).
% 1.98/2.16  all X Y Z (r1(X,Y)&r1(Y,Z)->r1(X,Z)).
% 1.98/2.16  -(-(exists X (-((all Y (-r1(X,Y)| -p3(Y)| (all X (-r1(Y,X)| -p1(X)))))| (all Y (-r1(X,Y)| -(all X (-r1(Y,X)| -((all Y (-r1(X,Y)))| -(all Y (-r1(X,Y)| -(p2(Y)& -p1(Y)| -p2(Y)&p1(Y))))| (all Y (-r1(X,Y)|p3(Y)))| (all Y (-r1(X,Y)| -p1(Y)& -p2(Y)|p2(Y)&p1(Y)))))))))))).
% 1.98/2.16  end_of_list.
% 1.98/2.16  
% 1.98/2.16  -------> usable clausifies to:
% 1.98/2.16  
% 1.98/2.16  list(usable).
% 1.98/2.16  0 [] r1(X,X).
% 1.98/2.16  0 [] -r1(X,Y)| -r1(Y,Z)|r1(X,Z).
% 1.98/2.16  0 [] r1($c4,$c2).
% 1.98/2.16  0 [] p3($c2).
% 1.98/2.16  0 [] r1($c2,$c1).
% 1.98/2.16  0 [] p1($c1).
% 1.98/2.16  0 [] r1($c4,$c3).
% 1.98/2.16  0 [] -r1($c3,X)|r1(X,$f1(X)).
% 1.98/2.16  0 [] -r1($c3,X)| -r1(X,Y)| -p2(Y)|p1(Y).
% 1.98/2.16  0 [] -r1($c3,X)| -r1(X,Y)|p2(Y)| -p1(Y).
% 1.98/2.16  0 [] -r1($c3,X)|r1(X,$f2(X)).
% 1.98/2.16  0 [] -r1($c3,X)| -p3($f2(X)).
% 1.98/2.16  0 [] -r1($c3,X)|r1(X,$f3(X)).
% 1.98/2.16  0 [] -r1($c3,X)|p1($f3(X))|p2($f3(X)).
% 1.98/2.16  0 [] -r1($c3,X)| -p2($f3(X))| -p1($f3(X)).
% 1.98/2.16  end_of_list.
% 1.98/2.16  
% 1.98/2.16  SCAN INPUT: prop=0, horn=0, equality=0, symmetry=0, max_lits=4.
% 1.98/2.16  
% 1.98/2.16  This is a non-Horn set without equality.  The strategy will
% 1.98/2.16  be ordered hyper_res, unit deletion, and factoring, with
% 1.98/2.16  satellites in sos and with nuclei in usable.
% 1.98/2.16  
% 1.98/2.16     dependent: set(hyper_res).
% 1.98/2.16     dependent: set(factor).
% 1.98/2.16     dependent: set(unit_deletion).
% 1.98/2.16  
% 1.98/2.16  ------------> process usable:
% 1.98/2.16  ** KEPT (pick-wt=9): 1 [] -r1(A,B)| -r1(B,C)|r1(A,C).
% 1.98/2.16  ** KEPT (pick-wt=7): 2 [] -r1($c3,A)|r1(A,$f1(A)).
% 1.98/2.16  ** KEPT (pick-wt=10): 3 [] -r1($c3,A)| -r1(A,B)| -p2(B)|p1(B).
% 1.98/2.16  ** KEPT (pick-wt=10): 4 [] -r1($c3,A)| -r1(A,B)|p2(B)| -p1(B).
% 1.98/2.16  ** KEPT (pick-wt=7): 5 [] -r1($c3,A)|r1(A,$f2(A)).
% 1.98/2.16  ** KEPT (pick-wt=6): 6 [] -r1($c3,A)| -p3($f2(A)).
% 1.98/2.16  ** KEPT (pick-wt=7): 7 [] -r1($c3,A)|r1(A,$f3(A)).
% 1.98/2.16  ** KEPT (pick-wt=9): 8 [] -r1($c3,A)|p1($f3(A))|p2($f3(A)).
% 1.98/2.16  ** KEPT (pick-wt=9): 9 [] -r1($c3,A)| -p2($f3(A))| -p1($f3(A)).
% 1.98/2.16  
% 1.98/2.16  ------------> process sos:
% 1.98/2.16  ** KEPT (pick-wt=3): 12 [] r1(A,A).
% 1.98/2.16  ** KEPT (pick-wt=3): 13 [] r1($c4,$c2).
% 1.98/2.16  ** KEPT (pick-wt=2): 14 [] p3($c2).
% 1.98/2.16  ** KEPT (pick-wt=3): 15 [] r1($c2,$c1).
% 1.98/2.16  ** KEPT (pick-wt=2): 16 [] p1($c1).
% 1.98/2.16  ** KEPT (pick-wt=3): 17 [] r1($c4,$c3).
% 1.98/2.16  
% 1.98/2.16  ======= end of input processing =======
% 1.98/2.16  
% 1.98/2.16  =========== start of search ===========
% 1.98/2.16  
% 1.98/2.16  -------- PROOF -------- 
% 1.98/2.16  
% 1.98/2.16  -----> EMPTY CLAUSE at   0.00 sec ----> 40 [hyper,39,9,12,38] $F.
% 1.98/2.16  
% 1.98/2.16  Length of proof is 4.  Level of proof is 3.
% 1.98/2.16  
% 1.98/2.16  ---------------- PROOF ----------------
% 1.98/2.16  % SZS status Theorem
% 1.98/2.16  % SZS output start Refutation
% See solution above
% 1.98/2.16  ------------ end of proof -------------
% 1.98/2.16  
% 1.98/2.16  
% 1.98/2.16  Search stopped by max_proofs option.
% 1.98/2.16  
% 1.98/2.16  
% 1.98/2.16  Search stopped by max_proofs option.
% 1.98/2.16  
% 1.98/2.16  ============ end of search ============
% 1.98/2.16  
% 1.98/2.16  -------------- statistics -------------
% 1.98/2.16  clauses given                 13
% 1.98/2.16  clauses generated             44
% 1.98/2.16  clauses kept                  39
% 1.98/2.16  clauses forward subsumed      18
% 1.98/2.16  clauses back subsumed          1
% 1.98/2.16  Kbytes malloced              976
% 1.98/2.16  
% 1.98/2.16  ----------- times (seconds) -----------
% 1.98/2.16  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.98/2.16  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.98/2.16  wall-clock time        2             (0 hr, 0 min, 2 sec)
% 1.98/2.16  
% 1.98/2.16  That finishes the proof of the theorem.
% 1.98/2.16  
% 1.98/2.16  Process 18840 finished Wed Jul 27 09:29:12 2022
% 1.98/2.16  Otter interrupted
% 1.98/2.16  PROOF FOUND
%------------------------------------------------------------------------------