TSTP Solution File: NUM531+2 by Otter---3.3

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Otter---3.3
% Problem  : NUM531+2 : TPTP v8.1.0. Released v4.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : otter-tptp-script %s

% Computer : n017.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 13:08:44 EDT 2022

% Result   : Theorem 1.71s 1.92s
% Output   : Refutation 1.71s
% Verified : 
% SZS Type : Refutation
%            Derivation depth      :    3
%            Number of leaves      :    5
% Syntax   : Number of clauses     :    8 (   7 unt;   0 nHn;   8 RR)
%            Number of literals    :   10 (   2 equ;   3 neg)
%            Maximal clause size   :    3 (   1 avg)
%            Maximal term depth    :    1 (   1 avg)
%            Number of predicates  :    5 (   3 usr;   1 prp; 0-2 aty)
%            Number of functors    :    2 (   2 usr;   2 con; 0-0 aty)
%            Number of variables   :    1 (   0 sgn)

% Comments : 
%------------------------------------------------------------------------------
cnf(5,axiom,
    ( ~ aSet0(A)
    | ~ isCountable0(A)
    | ~ isFinite0(A) ),
    file('NUM531+2.p',unknown),
    [] ).

cnf(8,axiom,
    isFinite0(slcrc0),
    file('NUM531+2.p',unknown),
    [] ).

cnf(9,axiom,
    aSet0(dollar_c1),
    file('NUM531+2.p',unknown),
    [] ).

cnf(10,axiom,
    isCountable0(dollar_c1),
    file('NUM531+2.p',unknown),
    [] ).

cnf(11,axiom,
    dollar_c1 = slcrc0,
    file('NUM531+2.p',unknown),
    [] ).

cnf(13,plain,
    slcrc0 = dollar_c1,
    inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[11])]),
    [iquote('copy,11,flip.1')] ).

cnf(14,plain,
    isFinite0(dollar_c1),
    inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[8]),13]),
    [iquote('back_demod,8,demod,13')] ).

cnf(18,plain,
    $false,
    inference(hyper,[status(thm)],[14,5,9,10]),
    [iquote('hyper,14,5,9,10')] ).

%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.10/0.11  % Problem  : NUM531+2 : TPTP v8.1.0. Released v4.0.0.
% 0.10/0.12  % Command  : otter-tptp-script %s
% 0.12/0.33  % Computer : n017.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 09:15:33 EDT 2022
% 0.12/0.33  % CPUTime  : 
% 1.71/1.92  ----- Otter 3.3f, August 2004 -----
% 1.71/1.92  The process was started by sandbox on n017.cluster.edu,
% 1.71/1.92  Wed Jul 27 09:15:33 2022
% 1.71/1.92  The command was "./otter".  The process ID is 19772.
% 1.71/1.92  
% 1.71/1.92  set(prolog_style_variables).
% 1.71/1.92  set(auto).
% 1.71/1.92     dependent: set(auto1).
% 1.71/1.92     dependent: set(process_input).
% 1.71/1.92     dependent: clear(print_kept).
% 1.71/1.92     dependent: clear(print_new_demod).
% 1.71/1.92     dependent: clear(print_back_demod).
% 1.71/1.92     dependent: clear(print_back_sub).
% 1.71/1.92     dependent: set(control_memory).
% 1.71/1.92     dependent: assign(max_mem, 12000).
% 1.71/1.92     dependent: assign(pick_given_ratio, 4).
% 1.71/1.92     dependent: assign(stats_level, 1).
% 1.71/1.92     dependent: assign(max_seconds, 10800).
% 1.71/1.92  clear(print_given).
% 1.71/1.92  
% 1.71/1.92  formula_list(usable).
% 1.71/1.92  all A (A=A).
% 1.71/1.92  all W0 (aSet0(W0)->$T).
% 1.71/1.92  all W0 (aElement0(W0)->$T).
% 1.71/1.92  all W0 (aSet0(W0)-> (all W1 (aElementOf0(W1,W0)->aElement0(W1)))).
% 1.71/1.92  all W0 (aSet0(W0)-> (isFinite0(W0)->$T)).
% 1.71/1.92  all W0 (W0=slcrc0<->aSet0(W0)& -(exists W1 aElementOf0(W1,W0))).
% 1.71/1.92  isFinite0(slcrc0).
% 1.71/1.92  all W0 (aSet0(W0)-> (isCountable0(W0)->$T)).
% 1.71/1.92  all W0 (aSet0(W0)&isCountable0(W0)-> -isFinite0(W0)).
% 1.71/1.92  -(all W0 (aSet0(W0)&isCountable0(W0)-> -(-(exists W1 aElementOf0(W1,W0))&W0=slcrc0))).
% 1.71/1.92  end_of_list.
% 1.71/1.92  
% 1.71/1.92  -------> usable clausifies to:
% 1.71/1.92  
% 1.71/1.92  list(usable).
% 1.71/1.92  0 [] A=A.
% 1.71/1.92  0 [] -aSet0(W0)|$T.
% 1.71/1.92  0 [] -aElement0(W0)|$T.
% 1.71/1.92  0 [] -aSet0(W0)| -aElementOf0(W1,W0)|aElement0(W1).
% 1.71/1.92  0 [] -aSet0(W0)| -isFinite0(W0)|$T.
% 1.71/1.92  0 [] W0!=slcrc0|aSet0(W0).
% 1.71/1.92  0 [] W0!=slcrc0| -aElementOf0(W1,W0).
% 1.71/1.92  0 [] W0=slcrc0| -aSet0(W0)|aElementOf0($f1(W0),W0).
% 1.71/1.92  0 [] isFinite0(slcrc0).
% 1.71/1.92  0 [] -aSet0(W0)| -isCountable0(W0)|$T.
% 1.71/1.92  0 [] -aSet0(W0)| -isCountable0(W0)| -isFinite0(W0).
% 1.71/1.92  0 [] aSet0($c1).
% 1.71/1.92  0 [] isCountable0($c1).
% 1.71/1.92  0 [] -aElementOf0(W1,$c1).
% 1.71/1.92  0 [] $c1=slcrc0.
% 1.71/1.92  end_of_list.
% 1.71/1.92  
% 1.71/1.92  SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.71/1.92  
% 1.71/1.92  This ia a non-Horn set with equality.  The strategy will be
% 1.71/1.92  Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.71/1.92  deletion, with positive clauses in sos and nonpositive
% 1.71/1.92  clauses in usable.
% 1.71/1.92  
% 1.71/1.92     dependent: set(knuth_bendix).
% 1.71/1.92     dependent: set(anl_eq).
% 1.71/1.92     dependent: set(para_from).
% 1.71/1.92     dependent: set(para_into).
% 1.71/1.92     dependent: clear(para_from_right).
% 1.71/1.92     dependent: clear(para_into_right).
% 1.71/1.92     dependent: set(para_from_vars).
% 1.71/1.92     dependent: set(eq_units_both_ways).
% 1.71/1.92     dependent: set(dynamic_demod_all).
% 1.71/1.92     dependent: set(dynamic_demod).
% 1.71/1.92     dependent: set(order_eq).
% 1.71/1.92     dependent: set(back_demod).
% 1.71/1.92     dependent: set(lrpo).
% 1.71/1.92     dependent: set(hyper_res).
% 1.71/1.92     dependent: set(unit_deletion).
% 1.71/1.92     dependent: set(factor).
% 1.71/1.92  
% 1.71/1.92  ------------> process usable:
% 1.71/1.92  ** KEPT (pick-wt=7): 1 [] -aSet0(A)| -aElementOf0(B,A)|aElement0(B).
% 1.71/1.92  ** KEPT (pick-wt=5): 2 [] A!=slcrc0|aSet0(A).
% 1.71/1.92  ** KEPT (pick-wt=6): 3 [] A!=slcrc0| -aElementOf0(B,A).
% 1.71/1.92  ** KEPT (pick-wt=9): 4 [] A=slcrc0| -aSet0(A)|aElementOf0($f1(A),A).
% 1.71/1.92  ** KEPT (pick-wt=6): 5 [] -aSet0(A)| -isCountable0(A)| -isFinite0(A).
% 1.71/1.92  ** KEPT (pick-wt=3): 6 [] -aElementOf0(A,$c1).
% 1.71/1.92  
% 1.71/1.92  ------------> process sos:
% 1.71/1.92  ** KEPT (pick-wt=3): 7 [] A=A.
% 1.71/1.92  ** KEPT (pick-wt=2): 8 [] isFinite0(slcrc0).
% 1.71/1.92  ** KEPT (pick-wt=2): 9 [] aSet0($c1).
% 1.71/1.92  ** KEPT (pick-wt=2): 10 [] isCountable0($c1).
% 1.71/1.92  ** KEPT (pick-wt=3): 12 [copy,11,flip.1] slcrc0=$c1.
% 1.71/1.92  ---> New Demodulator: 13 [new_demod,12] slcrc0=$c1.
% 1.71/1.92    Following clause subsumed by 7 during input processing: 0 [copy,7,flip.1] A=A.
% 1.71/1.92  >>>> Starting back demodulation with 13.
% 1.71/1.92      >> back demodulating 8 with 13.
% 1.71/1.92      >> back demodulating 4 with 13.
% 1.71/1.92      >> back demodulating 3 with 13.
% 1.71/1.92      >> back demodulating 2 with 13.
% 1.71/1.92  
% 1.71/1.92  ======= end of input processing =======
% 1.71/1.92  
% 1.71/1.92  =========== start of search ===========
% 1.71/1.92  
% 1.71/1.92  -------- PROOF -------- 
% 1.71/1.92  
% 1.71/1.92  -----> EMPTY CLAUSE at   0.00 sec ----> 18 [hyper,14,5,9,10] $F.
% 1.71/1.92  
% 1.71/1.92  Length of proof is 2.  Level of proof is 2.
% 1.71/1.92  
% 1.71/1.92  ---------------- PROOF ----------------
% 1.71/1.92  % SZS status Theorem
% 1.71/1.92  % SZS output start Refutation
% See solution above
% 1.71/1.92  ------------ end of proof -------------
% 1.71/1.92  
% 1.71/1.92  
% 1.71/1.92  Search stopped by max_proofs option.
% 1.71/1.92  
% 1.71/1.92  
% 1.71/1.92  Search stopped by max_proofs option.
% 1.71/1.92  
% 1.71/1.92  ============ end of search ============
% 1.71/1.92  
% 1.71/1.92  -------------- statistics -------------
% 1.71/1.92  clauses given                  4
% 1.71/1.92  clauses generated              1
% 1.71/1.92  clauses kept                  15
% 1.71/1.92  clauses forward subsumed       1
% 1.71/1.92  clauses back subsumed          0
% 1.71/1.92  Kbytes malloced              976
% 1.71/1.92  
% 1.71/1.92  ----------- times (seconds) -----------
% 1.71/1.92  user CPU time          0.00          (0 hr, 0 min, 0 sec)
% 1.71/1.92  system CPU time        0.00          (0 hr, 0 min, 0 sec)
% 1.71/1.92  wall-clock time        1             (0 hr, 0 min, 1 sec)
% 1.71/1.92  
% 1.71/1.92  That finishes the proof of the theorem.
% 1.71/1.92  
% 1.71/1.92  Process 19772 finished Wed Jul 27 09:15:34 2022
% 1.71/1.92  Otter interrupted
% 1.71/1.92  PROOF FOUND
%------------------------------------------------------------------------------