TSTP Solution File: SYN411+1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SYN411+1 : TPTP v8.1.0. Released v2.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n005.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 13:24:29 EDT 2022
% Result : Theorem 1.62s 1.80s
% Output : Refutation 1.62s
% Verified :
% SZS Type : Refutation
% Derivation depth : 2
% Number of leaves : 2
% Syntax : Number of clauses : 4 ( 2 unt; 1 nHn; 2 RR)
% Number of literals : 6 ( 0 equ; 2 neg)
% Maximal clause size : 2 ( 1 avg)
% Maximal term depth : 1 ( 1 avg)
% Number of predicates : 2 ( 1 usr; 1 prp; 0-3 aty)
% Number of functors : 6 ( 6 usr; 6 con; 0-0 aty)
% Number of variables : 9 ( 9 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( ~ f(dollar_c3,dollar_c2,dollar_c1)
| ~ f(dollar_c6,dollar_c5,dollar_c4) ),
file('SYN411+1.p',unknown),
[] ).
cnf(2,axiom,
( f(A,B,C)
| f(D,E,F) ),
file('SYN411+1.p',unknown),
[] ).
cnf(3,plain,
f(A,B,C),
inference(factor_simp,[status(thm)],[inference(copy,[status(thm)],[2])]),
[iquote('copy,2,factor_simp')] ).
cnf(4,plain,
$false,
inference(hyper,[status(thm)],[3,1,3]),
[iquote('hyper,3,1,3')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : SYN411+1 : TPTP v8.1.0. Released v2.0.0.
% 0.11/0.13 % Command : otter-tptp-script %s
% 0.13/0.34 % Computer : n005.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Wed Jul 27 11:12:35 EDT 2022
% 0.13/0.34 % CPUTime :
% 1.62/1.80 ----- Otter 3.3f, August 2004 -----
% 1.62/1.80 The process was started by sandbox2 on n005.cluster.edu,
% 1.62/1.80 Wed Jul 27 11:12:35 2022
% 1.62/1.80 The command was "./otter". The process ID is 5008.
% 1.62/1.80
% 1.62/1.80 set(prolog_style_variables).
% 1.62/1.80 set(auto).
% 1.62/1.80 dependent: set(auto1).
% 1.62/1.80 dependent: set(process_input).
% 1.62/1.80 dependent: clear(print_kept).
% 1.62/1.80 dependent: clear(print_new_demod).
% 1.62/1.80 dependent: clear(print_back_demod).
% 1.62/1.80 dependent: clear(print_back_sub).
% 1.62/1.80 dependent: set(control_memory).
% 1.62/1.80 dependent: assign(max_mem, 12000).
% 1.62/1.80 dependent: assign(pick_given_ratio, 4).
% 1.62/1.80 dependent: assign(stats_level, 1).
% 1.62/1.80 dependent: assign(max_seconds, 10800).
% 1.62/1.80 clear(print_given).
% 1.62/1.80
% 1.62/1.80 formula_list(usable).
% 1.62/1.80 -((all X Y Z f(X,Y,Z))<-> -(exists U V W (-f(U,V,W)))).
% 1.62/1.80 end_of_list.
% 1.62/1.80
% 1.62/1.80 -------> usable clausifies to:
% 1.62/1.80
% 1.62/1.80 list(usable).
% 1.62/1.80 0 [] f(X,Y,Z)|f(U,V,W).
% 1.62/1.80 0 [] -f($c3,$c2,$c1)| -f($c6,$c5,$c4).
% 1.62/1.80 end_of_list.
% 1.62/1.80
% 1.62/1.80 SCAN INPUT: prop=0, horn=0, equality=0, symmetry=0, max_lits=2.
% 1.62/1.80
% 1.62/1.80 This is a non-Horn set without equality. The strategy will
% 1.62/1.80 be ordered hyper_res, unit deletion, and factoring, with
% 1.62/1.80 satellites in sos and with nuclei in usable.
% 1.62/1.80
% 1.62/1.80 dependent: set(hyper_res).
% 1.62/1.80 dependent: set(factor).
% 1.62/1.80 dependent: set(unit_deletion).
% 1.62/1.80
% 1.62/1.80 ------------> process usable:
% 1.62/1.80 ** KEPT (pick-wt=8): 1 [] -f($c3,$c2,$c1)| -f($c6,$c5,$c4).
% 1.62/1.80
% 1.62/1.80 ------------> process sos:
% 1.62/1.80 ** KEPT (pick-wt=4): 3 [copy,2,factor_simp] f(A,B,C).
% 1.62/1.80
% 1.62/1.80 ======= end of input processing =======
% 1.62/1.80
% 1.62/1.80 =========== start of search ===========
% 1.62/1.80
% 1.62/1.80 -------- PROOF --------
% 1.62/1.80
% 1.62/1.80 -----> EMPTY CLAUSE at 0.00 sec ----> 4 [hyper,3,1,3] $F.
% 1.62/1.80
% 1.62/1.80 Length of proof is 1. Level of proof is 1.
% 1.62/1.80
% 1.62/1.80 ---------------- PROOF ----------------
% 1.62/1.80 % SZS status Theorem
% 1.62/1.80 % SZS output start Refutation
% See solution above
% 1.62/1.80 ------------ end of proof -------------
% 1.62/1.80
% 1.62/1.80
% 1.62/1.80 Search stopped by max_proofs option.
% 1.62/1.80
% 1.62/1.80
% 1.62/1.80 Search stopped by max_proofs option.
% 1.62/1.80
% 1.62/1.80 ============ end of search ============
% 1.62/1.80
% 1.62/1.80 -------------- statistics -------------
% 1.62/1.80 clauses given 1
% 1.62/1.80 clauses generated 1
% 1.62/1.80 clauses kept 2
% 1.62/1.80 clauses forward subsumed 0
% 1.62/1.80 clauses back subsumed 0
% 1.62/1.80 Kbytes malloced 976
% 1.62/1.80
% 1.62/1.80 ----------- times (seconds) -----------
% 1.62/1.80 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.62/1.80 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.62/1.80 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.62/1.80
% 1.62/1.80 That finishes the proof of the theorem.
% 1.62/1.80
% 1.62/1.80 Process 5008 finished Wed Jul 27 11:12:37 2022
% 1.62/1.80 Otter interrupted
% 1.62/1.80 PROOF FOUND
%------------------------------------------------------------------------------