TSTP Solution File: HEN007-4 by Otter---3.3
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : HEN007-4 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n019.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 12:57:57 EDT 2022
% Result : Unsatisfiable 1.64s 1.83s
% Output : Refutation 1.64s
% Verified :
% SZS Type : Refutation
% Derivation depth : 5
% Number of leaves : 6
% Syntax : Number of clauses : 11 ( 8 unt; 0 nHn; 6 RR)
% Number of literals : 15 ( 2 equ; 5 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 3 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 5 ( 5 usr; 4 con; 0-2 aty)
% Number of variables : 14 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(2,axiom,
( divide(A,B) != zero
| less_e_qual(A,B) ),
file('HEN007-4.p',unknown),
[] ).
cnf(4,axiom,
( ~ less_e_qual(A,B)
| ~ less_e_qual(B,C)
| less_e_qual(A,C) ),
file('HEN007-4.p',unknown),
[] ).
cnf(5,axiom,
( ~ less_e_qual(divide(A,B),C)
| less_e_qual(divide(A,C),B) ),
file('HEN007-4.p',unknown),
[] ).
cnf(6,axiom,
~ less_e_qual(divide(c,b),divide(c,a)),
file('HEN007-4.p',unknown),
[] ).
cnf(16,axiom,
divide(A,A) = zero,
file('HEN007-4.p',unknown),
[] ).
cnf(20,axiom,
less_e_qual(a,b),
file('HEN007-4.p',unknown),
[] ).
cnf(50,plain,
less_e_qual(A,A),
inference(hyper,[status(thm)],[16,2]),
[iquote('hyper,16,2')] ).
cnf(53,plain,
less_e_qual(divide(A,divide(A,B)),B),
inference(hyper,[status(thm)],[50,5]),
[iquote('hyper,50,5')] ).
cnf(105,plain,
less_e_qual(divide(A,divide(A,a)),b),
inference(hyper,[status(thm)],[53,4,20]),
[iquote('hyper,53,4,20')] ).
cnf(129,plain,
less_e_qual(divide(A,b),divide(A,a)),
inference(hyper,[status(thm)],[105,5]),
[iquote('hyper,105,5')] ).
cnf(130,plain,
$false,
inference(binary,[status(thm)],[129,6]),
[iquote('binary,129.1,6.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12 % Problem : HEN007-4 : TPTP v8.1.0. Released v1.0.0.
% 0.04/0.13 % Command : otter-tptp-script %s
% 0.14/0.34 % Computer : n019.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Wed Jul 27 09:00:52 EDT 2022
% 0.14/0.34 % CPUTime :
% 1.64/1.83 ----- Otter 3.3f, August 2004 -----
% 1.64/1.83 The process was started by sandbox2 on n019.cluster.edu,
% 1.64/1.83 Wed Jul 27 09:00:52 2022
% 1.64/1.83 The command was "./otter". The process ID is 4364.
% 1.64/1.83
% 1.64/1.83 set(prolog_style_variables).
% 1.64/1.83 set(auto).
% 1.64/1.83 dependent: set(auto1).
% 1.64/1.83 dependent: set(process_input).
% 1.64/1.83 dependent: clear(print_kept).
% 1.64/1.83 dependent: clear(print_new_demod).
% 1.64/1.83 dependent: clear(print_back_demod).
% 1.64/1.83 dependent: clear(print_back_sub).
% 1.64/1.83 dependent: set(control_memory).
% 1.64/1.83 dependent: assign(max_mem, 12000).
% 1.64/1.83 dependent: assign(pick_given_ratio, 4).
% 1.64/1.83 dependent: assign(stats_level, 1).
% 1.64/1.83 dependent: assign(max_seconds, 10800).
% 1.64/1.83 clear(print_given).
% 1.64/1.83
% 1.64/1.83 list(usable).
% 1.64/1.83 0 [] A=A.
% 1.64/1.83 0 [] -less_e_qual(X,Y)|divide(X,Y)=zero.
% 1.64/1.83 0 [] divide(X,Y)!=zero|less_e_qual(X,Y).
% 1.64/1.83 0 [] less_e_qual(divide(X,Y),X).
% 1.64/1.83 0 [] less_e_qual(divide(divide(X,Z),divide(Y,Z)),divide(divide(X,Y),Z)).
% 1.64/1.83 0 [] less_e_qual(zero,X).
% 1.64/1.83 0 [] -less_e_qual(X,Y)| -less_e_qual(Y,X)|X=Y.
% 1.64/1.83 0 [] less_e_qual(X,identity).
% 1.64/1.83 0 [] divide(X,identity)=zero.
% 1.64/1.83 0 [] divide(zero,X)=zero.
% 1.64/1.83 0 [] divide(X,X)=zero.
% 1.64/1.83 0 [] divide(a,zero)=a.
% 1.64/1.83 0 [] -less_e_qual(X,Y)| -less_e_qual(Y,Z)|less_e_qual(X,Z).
% 1.64/1.83 0 [] -less_e_qual(divide(X,Y),Z)|less_e_qual(divide(X,Z),Y).
% 1.64/1.83 0 [] less_e_qual(a,b).
% 1.64/1.83 0 [] -less_e_qual(divide(c,b),divide(c,a)).
% 1.64/1.83 end_of_list.
% 1.64/1.83
% 1.64/1.83 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=3.
% 1.64/1.83
% 1.64/1.83 This is a Horn set with equality. The strategy will be
% 1.64/1.83 Knuth-Bendix and hyper_res, with positive clauses in
% 1.64/1.83 sos and nonpositive clauses in usable.
% 1.64/1.83
% 1.64/1.83 dependent: set(knuth_bendix).
% 1.64/1.83 dependent: set(anl_eq).
% 1.64/1.83 dependent: set(para_from).
% 1.64/1.83 dependent: set(para_into).
% 1.64/1.83 dependent: clear(para_from_right).
% 1.64/1.83 dependent: clear(para_into_right).
% 1.64/1.83 dependent: set(para_from_vars).
% 1.64/1.83 dependent: set(eq_units_both_ways).
% 1.64/1.83 dependent: set(dynamic_demod_all).
% 1.64/1.83 dependent: set(dynamic_demod).
% 1.64/1.83 dependent: set(order_eq).
% 1.64/1.83 dependent: set(back_demod).
% 1.64/1.83 dependent: set(lrpo).
% 1.64/1.83 dependent: set(hyper_res).
% 1.64/1.83 dependent: clear(order_hyper).
% 1.64/1.83
% 1.64/1.83 ------------> process usable:
% 1.64/1.83 ** KEPT (pick-wt=8): 1 [] -less_e_qual(A,B)|divide(A,B)=zero.
% 1.64/1.83 ** KEPT (pick-wt=8): 2 [] divide(A,B)!=zero|less_e_qual(A,B).
% 1.64/1.83 ** KEPT (pick-wt=9): 3 [] -less_e_qual(A,B)| -less_e_qual(B,A)|A=B.
% 1.64/1.83 ** KEPT (pick-wt=9): 4 [] -less_e_qual(A,B)| -less_e_qual(B,C)|less_e_qual(A,C).
% 1.64/1.83 ** KEPT (pick-wt=10): 5 [] -less_e_qual(divide(A,B),C)|less_e_qual(divide(A,C),B).
% 1.64/1.83 ** KEPT (pick-wt=7): 6 [] -less_e_qual(divide(c,b),divide(c,a)).
% 1.64/1.83
% 1.64/1.83 ------------> process sos:
% 1.64/1.83 ** KEPT (pick-wt=3): 7 [] A=A.
% 1.64/1.83 ** KEPT (pick-wt=5): 8 [] less_e_qual(divide(A,B),A).
% 1.64/1.83 ** KEPT (pick-wt=13): 9 [] less_e_qual(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B)).
% 1.64/1.83 ** KEPT (pick-wt=3): 10 [] less_e_qual(zero,A).
% 1.64/1.83 ** KEPT (pick-wt=3): 11 [] less_e_qual(A,identity).
% 1.64/1.83 ** KEPT (pick-wt=5): 12 [] divide(A,identity)=zero.
% 1.64/1.83 ---> New Demodulator: 13 [new_demod,12] divide(A,identity)=zero.
% 1.64/1.83 ** KEPT (pick-wt=5): 14 [] divide(zero,A)=zero.
% 1.64/1.83 ---> New Demodulator: 15 [new_demod,14] divide(zero,A)=zero.
% 1.64/1.83 ** KEPT (pick-wt=5): 16 [] divide(A,A)=zero.
% 1.64/1.83 ---> New Demodulator: 17 [new_demod,16] divide(A,A)=zero.
% 1.64/1.83 ** KEPT (pick-wt=5): 18 [] divide(a,zero)=a.
% 1.64/1.83 ---> New Demodulator: 19 [new_demod,18] divide(a,zero)=a.
% 1.64/1.83 ** KEPT (pick-wt=3): 20 [] less_e_qual(a,b).
% 1.64/1.83 Following clause subsumed by 7 during input processing: 0 [copy,7,flip.1] A=A.
% 1.64/1.83 >>>> Starting back demodulation with 13.
% 1.64/1.83 >>>> Starting back demodulation with 15.
% 1.64/1.83 >>>> Starting back demodulation with 17.
% 1.64/1.83 >>>> Starting back demodulation with 19.
% 1.64/1.83
% 1.64/1.83 ======= end of input processing =======
% 1.64/1.83
% 1.64/1.83 =========== start of search ===========
% 1.64/1.83
% 1.64/1.83 -------- PROOF --------
% 1.64/1.83
% 1.64/1.83 ----> UNIT CONFLICT at 0.01 sec ----> 130 [binary,129.1,6.1] $F.
% 1.64/1.83
% 1.64/1.83 Length of proof is 4. Level of proof is 4.
% 1.64/1.83
% 1.64/1.83 ---------------- PROOF ----------------
% 1.64/1.83 % SZS status Unsatisfiable
% 1.64/1.83 % SZS output start Refutation
% See solution above
% 1.64/1.83 ------------ end of proof -------------
% 1.64/1.83
% 1.64/1.83
% 1.64/1.83 Search stopped by max_proofs option.
% 1.64/1.83
% 1.64/1.83
% 1.64/1.83 Search stopped by max_proofs option.
% 1.64/1.83
% 1.64/1.83 ============ end of search ============
% 1.64/1.83
% 1.64/1.83 -------------- statistics -------------
% 1.64/1.83 clauses given 24
% 1.64/1.83 clauses generated 336
% 1.64/1.83 clauses kept 117
% 1.64/1.83 clauses forward subsumed 236
% 1.64/1.83 clauses back subsumed 0
% 1.64/1.83 Kbytes malloced 976
% 1.64/1.83
% 1.64/1.83 ----------- times (seconds) -----------
% 1.64/1.83 user CPU time 0.01 (0 hr, 0 min, 0 sec)
% 1.64/1.83 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.64/1.83 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.64/1.83
% 1.64/1.83 That finishes the proof of the theorem.
% 1.64/1.83
% 1.64/1.83 Process 4364 finished Wed Jul 27 09:00:53 2022
% 1.64/1.83 Otter interrupted
% 1.64/1.83 PROOF FOUND
%------------------------------------------------------------------------------