TSTP Solution File: SYN939+1 by Otter---3.3
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SYN939+1 : TPTP v8.1.0. Released v3.1.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n018.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 13:25:31 EDT 2022
% Result : Theorem 1.68s 1.88s
% Output : Refutation 1.68s
% Verified :
% SZS Type : Refutation
% Derivation depth : 6
% Number of leaves : 5
% Syntax : Number of clauses : 11 ( 4 unt; 4 nHn; 4 RR)
% Number of literals : 24 ( 0 equ; 10 neg)
% Maximal clause size : 4 ( 2 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 4 ( 3 usr; 1 prp; 0-1 aty)
% Number of functors : 3 ( 3 usr; 2 con; 0-1 aty)
% Number of variables : 11 ( 7 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( p(f(A))
| r(A)
| ~ q(B) ),
file('SYN939+1.p',unknown),
[] ).
cnf(2,axiom,
( p(f(A))
| ~ r(dollar_c1)
| ~ r(dollar_c2)
| ~ q(B) ),
file('SYN939+1.p',unknown),
[] ).
cnf(3,axiom,
( ~ p(A)
| r(B)
| ~ q(A) ),
file('SYN939+1.p',unknown),
[] ).
cnf(4,axiom,
( ~ p(A)
| ~ r(dollar_c1)
| ~ r(dollar_c2)
| ~ q(A) ),
file('SYN939+1.p',unknown),
[] ).
cnf(5,axiom,
q(f(A)),
file('SYN939+1.p',unknown),
[] ).
cnf(6,plain,
( p(f(A))
| r(A) ),
inference(hyper,[status(thm)],[5,1]),
[iquote('hyper,5,1')] ).
cnf(7,plain,
( p(f(dollar_c2))
| p(f(dollar_c1)) ),
inference(factor_simp,[status(thm)],[inference(hyper,[status(thm)],[6,2,6,5])]),
[iquote('hyper,6,2,6,5,factor_simp')] ).
cnf(8,plain,
( p(f(dollar_c1))
| r(A) ),
inference(hyper,[status(thm)],[7,3,5]),
[iquote('hyper,7,3,5')] ).
cnf(10,plain,
p(f(dollar_c1)),
inference(factor_simp,[status(thm)],[inference(factor_simp,[status(thm)],[inference(hyper,[status(thm)],[8,4,7,8,5])])]),
[iquote('hyper,8,4,7,8,5,factor_simp,factor_simp')] ).
cnf(11,plain,
r(A),
inference(hyper,[status(thm)],[10,3,5]),
[iquote('hyper,10,3,5')] ).
cnf(12,plain,
$false,
inference(hyper,[status(thm)],[11,4,10,11,5]),
[iquote('hyper,11,4,10,11,5')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12 % Problem : SYN939+1 : TPTP v8.1.0. Released v3.1.0.
% 0.03/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n018.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Wed Jul 27 11:26:46 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.68/1.88 ----- Otter 3.3f, August 2004 -----
% 1.68/1.88 The process was started by sandbox2 on n018.cluster.edu,
% 1.68/1.88 Wed Jul 27 11:26:46 2022
% 1.68/1.88 The command was "./otter". The process ID is 15399.
% 1.68/1.88
% 1.68/1.88 set(prolog_style_variables).
% 1.68/1.88 set(auto).
% 1.68/1.88 dependent: set(auto1).
% 1.68/1.88 dependent: set(process_input).
% 1.68/1.88 dependent: clear(print_kept).
% 1.68/1.88 dependent: clear(print_new_demod).
% 1.68/1.88 dependent: clear(print_back_demod).
% 1.68/1.88 dependent: clear(print_back_sub).
% 1.68/1.88 dependent: set(control_memory).
% 1.68/1.88 dependent: assign(max_mem, 12000).
% 1.68/1.88 dependent: assign(pick_given_ratio, 4).
% 1.68/1.88 dependent: assign(stats_level, 1).
% 1.68/1.88 dependent: assign(max_seconds, 10800).
% 1.68/1.88 clear(print_given).
% 1.68/1.88
% 1.68/1.88 formula_list(usable).
% 1.68/1.88 -(all C B ((all Z q(f(Z)))-> (exists X Y ((p(f(Y))->p(X))& (r(Y)->r(B)&r(C))&q(X))))).
% 1.68/1.88 end_of_list.
% 1.68/1.88
% 1.68/1.88 -------> usable clausifies to:
% 1.68/1.88
% 1.68/1.88 list(usable).
% 1.68/1.88 0 [] q(f(Z)).
% 1.68/1.88 0 [] p(f(Y))|r(Y)| -q(X).
% 1.68/1.88 0 [] p(f(Y))| -r($c1)| -r($c2)| -q(X).
% 1.68/1.88 0 [] -p(X)|r(Y)| -q(X).
% 1.68/1.88 0 [] -p(X)| -r($c1)| -r($c2)| -q(X).
% 1.68/1.88 end_of_list.
% 1.68/1.88
% 1.68/1.88 SCAN INPUT: prop=0, horn=0, equality=0, symmetry=0, max_lits=4.
% 1.68/1.88
% 1.68/1.88 This is a non-Horn set without equality. The strategy will
% 1.68/1.88 be ordered hyper_res, unit deletion, and factoring, with
% 1.68/1.88 satellites in sos and with nuclei in usable.
% 1.68/1.88
% 1.68/1.88 dependent: set(hyper_res).
% 1.68/1.88 dependent: set(factor).
% 1.68/1.88 dependent: set(unit_deletion).
% 1.68/1.88
% 1.68/1.88 ------------> process usable:
% 1.68/1.88 ** KEPT (pick-wt=7): 1 [] p(f(A))|r(A)| -q(B).
% 1.68/1.88 ** KEPT (pick-wt=9): 2 [] p(f(A))| -r($c1)| -r($c2)| -q(B).
% 1.68/1.88 ** KEPT (pick-wt=6): 3 [] -p(A)|r(B)| -q(A).
% 1.68/1.88 ** KEPT (pick-wt=8): 4 [] -p(A)| -r($c1)| -r($c2)| -q(A).
% 1.68/1.88
% 1.68/1.88 ------------> process sos:
% 1.68/1.88 ** KEPT (pick-wt=3): 5 [] q(f(A)).
% 1.68/1.88
% 1.68/1.88 ======= end of input processing =======
% 1.68/1.88
% 1.68/1.88 =========== start of search ===========
% 1.68/1.88
% 1.68/1.88 -------- PROOF --------
% 1.68/1.88
% 1.68/1.88 -----> EMPTY CLAUSE at 0.00 sec ----> 12 [hyper,11,4,10,11,5] $F.
% 1.68/1.88
% 1.68/1.88 Length of proof is 5. Level of proof is 5.
% 1.68/1.88
% 1.68/1.88 ---------------- PROOF ----------------
% 1.68/1.88 % SZS status Theorem
% 1.68/1.88 % SZS output start Refutation
% See solution above
% 1.68/1.88 ------------ end of proof -------------
% 1.68/1.88
% 1.68/1.88
% 1.68/1.88 Search stopped by max_proofs option.
% 1.68/1.88
% 1.68/1.88
% 1.68/1.88 Search stopped by max_proofs option.
% 1.68/1.88
% 1.68/1.88 ============ end of search ============
% 1.68/1.88
% 1.68/1.88 -------------- statistics -------------
% 1.68/1.88 clauses given 6
% 1.68/1.88 clauses generated 22
% 1.68/1.88 clauses kept 11
% 1.68/1.88 clauses forward subsumed 15
% 1.68/1.88 clauses back subsumed 6
% 1.68/1.88 Kbytes malloced 976
% 1.68/1.88
% 1.68/1.88 ----------- times (seconds) -----------
% 1.68/1.88 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.68/1.88 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.68/1.88 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.68/1.88
% 1.68/1.88 That finishes the proof of the theorem.
% 1.68/1.88
% 1.68/1.88 Process 15399 finished Wed Jul 27 11:26:48 2022
% 1.68/1.88 Otter interrupted
% 1.68/1.88 PROOF FOUND
%------------------------------------------------------------------------------