TSTP Solution File: HEN004-5 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : HEN004-5 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n023.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 12:57:55 EDT 2022
% Result : Unsatisfiable 1.80s 2.06s
% Output : Refutation 1.80s
% Verified :
% SZS Type : Refutation
% Derivation depth : 6
% Number of leaves : 7
% Syntax : Number of clauses : 16 ( 11 unt; 0 nHn; 7 RR)
% Number of literals : 25 ( 24 equ; 12 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 5 ( 1 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 3 ( 3 usr; 2 con; 0-2 aty)
% Number of variables : 18 ( 2 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( divide(A,B) != zero
| divide(B,A) != zero
| A = B ),
file('HEN004-5.p',unknown),
[] ).
cnf(2,axiom,
divide(a,zero) != a,
file('HEN004-5.p',unknown),
[] ).
cnf(3,axiom,
A = A,
file('HEN004-5.p',unknown),
[] ).
cnf(4,axiom,
divide(divide(A,B),A) = zero,
file('HEN004-5.p',unknown),
[] ).
cnf(6,axiom,
divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B)) = zero,
file('HEN004-5.p',unknown),
[] ).
cnf(8,axiom,
divide(zero,A) = zero,
file('HEN004-5.p',unknown),
[] ).
cnf(12,axiom,
divide(A,A) = zero,
file('HEN004-5.p',unknown),
[] ).
cnf(16,plain,
( zero != zero
| divide(A,zero) != zero
| zero = A ),
inference(para_from,[status(thm),theory(equality)],[8,1]),
[iquote('para_from,8.1.1,1.1.1')] ).
cnf(21,plain,
( divide(A,divide(A,B)) != zero
| zero != zero
| divide(A,B) = A ),
inference(flip,[status(thm),theory(equality)],[inference(para_from,[status(thm),theory(equality)],[4,1])]),
[iquote('para_from,4.1.1,1.2.1,flip.3')] ).
cnf(25,plain,
divide(divide(divide(A,B),zero),divide(divide(A,zero),B)) = zero,
inference(para_into,[status(thm),theory(equality)],[6,8]),
[iquote('para_into,6.1.1.1.2,8.1.1')] ).
cnf(217,plain,
divide(divide(divide(A,divide(A,zero)),zero),zero) = zero,
inference(para_into,[status(thm),theory(equality)],[25,12]),
[iquote('para_into,25.1.1.2,12.1.1')] ).
cnf(269,plain,
( a != a
| divide(a,divide(a,zero)) != zero
| zero != zero ),
inference(para_from,[status(thm),theory(equality)],[21,2]),
[iquote('para_from,21.3.1,2.1.1')] ).
cnf(282,plain,
divide(divide(A,divide(A,zero)),zero) = zero,
inference(flip,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[217,16,3])]),
[iquote('hyper,217,16,3,flip.1')] ).
cnf(385,plain,
divide(A,divide(A,zero)) = zero,
inference(flip,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[282,16,3])]),
[iquote('hyper,282,16,3,flip.1')] ).
cnf(386,plain,
( a != a
| zero != zero ),
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[269]),385]),
[iquote('back_demod,269,demod,385')] ).
cnf(392,plain,
$false,
inference(hyper,[status(thm)],[386,3,3]),
[iquote('hyper,386,3,3')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.11 % Problem : HEN004-5 : TPTP v8.1.0. Released v1.0.0.
% 0.03/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n023.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 09:19:48 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.80/2.06 ----- Otter 3.3f, August 2004 -----
% 1.80/2.06 The process was started by sandbox on n023.cluster.edu,
% 1.80/2.06 Wed Jul 27 09:19:48 2022
% 1.80/2.06 The command was "./otter". The process ID is 32288.
% 1.80/2.06
% 1.80/2.06 set(prolog_style_variables).
% 1.80/2.06 set(auto).
% 1.80/2.06 dependent: set(auto1).
% 1.80/2.06 dependent: set(process_input).
% 1.80/2.06 dependent: clear(print_kept).
% 1.80/2.06 dependent: clear(print_new_demod).
% 1.80/2.06 dependent: clear(print_back_demod).
% 1.80/2.06 dependent: clear(print_back_sub).
% 1.80/2.06 dependent: set(control_memory).
% 1.80/2.06 dependent: assign(max_mem, 12000).
% 1.80/2.06 dependent: assign(pick_given_ratio, 4).
% 1.80/2.06 dependent: assign(stats_level, 1).
% 1.80/2.06 dependent: assign(max_seconds, 10800).
% 1.80/2.06 clear(print_given).
% 1.80/2.06
% 1.80/2.06 list(usable).
% 1.80/2.06 0 [] A=A.
% 1.80/2.06 0 [] divide(divide(X,Y),X)=zero.
% 1.80/2.06 0 [] divide(divide(divide(X,Z),divide(Y,Z)),divide(divide(X,Y),Z))=zero.
% 1.80/2.06 0 [] divide(zero,X)=zero.
% 1.80/2.06 0 [] divide(X,Y)!=zero|divide(Y,X)!=zero|X=Y.
% 1.80/2.06 0 [] divide(X,identity)=zero.
% 1.80/2.06 0 [] divide(X,X)=zero.
% 1.80/2.06 0 [] divide(a,zero)!=a.
% 1.80/2.06 end_of_list.
% 1.80/2.06
% 1.80/2.06 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=3.
% 1.80/2.06
% 1.80/2.06 This is a Horn set with equality. The strategy will be
% 1.80/2.06 Knuth-Bendix and hyper_res, with positive clauses in
% 1.80/2.06 sos and nonpositive clauses in usable.
% 1.80/2.06
% 1.80/2.06 dependent: set(knuth_bendix).
% 1.80/2.06 dependent: set(anl_eq).
% 1.80/2.06 dependent: set(para_from).
% 1.80/2.06 dependent: set(para_into).
% 1.80/2.06 dependent: clear(para_from_right).
% 1.80/2.06 dependent: clear(para_into_right).
% 1.80/2.06 dependent: set(para_from_vars).
% 1.80/2.06 dependent: set(eq_units_both_ways).
% 1.80/2.06 dependent: set(dynamic_demod_all).
% 1.80/2.06 dependent: set(dynamic_demod).
% 1.80/2.06 dependent: set(order_eq).
% 1.80/2.06 dependent: set(back_demod).
% 1.80/2.06 dependent: set(lrpo).
% 1.80/2.06 dependent: set(hyper_res).
% 1.80/2.06 dependent: clear(order_hyper).
% 1.80/2.06
% 1.80/2.06 ------------> process usable:
% 1.80/2.06 ** KEPT (pick-wt=13): 1 [] divide(A,B)!=zero|divide(B,A)!=zero|A=B.
% 1.80/2.06 ** KEPT (pick-wt=5): 2 [] divide(a,zero)!=a.
% 1.80/2.06
% 1.80/2.06 ------------> process sos:
% 1.80/2.06 ** KEPT (pick-wt=3): 3 [] A=A.
% 1.80/2.06 ** KEPT (pick-wt=7): 4 [] divide(divide(A,B),A)=zero.
% 1.80/2.06 ---> New Demodulator: 5 [new_demod,4] divide(divide(A,B),A)=zero.
% 1.80/2.06 ** KEPT (pick-wt=15): 6 [] divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B))=zero.
% 1.80/2.06 ---> New Demodulator: 7 [new_demod,6] divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B))=zero.
% 1.80/2.06 ** KEPT (pick-wt=5): 8 [] divide(zero,A)=zero.
% 1.80/2.06 ---> New Demodulator: 9 [new_demod,8] divide(zero,A)=zero.
% 1.80/2.06 ** KEPT (pick-wt=5): 10 [] divide(A,identity)=zero.
% 1.80/2.06 ---> New Demodulator: 11 [new_demod,10] divide(A,identity)=zero.
% 1.80/2.06 ** KEPT (pick-wt=5): 12 [] divide(A,A)=zero.
% 1.80/2.06 ---> New Demodulator: 13 [new_demod,12] divide(A,A)=zero.
% 1.80/2.06 Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.80/2.06 >>>> Starting back demodulation with 5.
% 1.80/2.06 >>>> Starting back demodulation with 7.
% 1.80/2.06 >>>> Starting back demodulation with 9.
% 1.80/2.06 >>>> Starting back demodulation with 11.
% 1.80/2.06 >>>> Starting back demodulation with 13.
% 1.80/2.06
% 1.80/2.06 ======= end of input processing =======
% 1.80/2.06
% 1.80/2.06 =========== start of search ===========
% 1.80/2.06
% 1.80/2.06 -------- PROOF --------
% 1.80/2.06
% 1.80/2.06 -----> EMPTY CLAUSE at 0.04 sec ----> 392 [hyper,386,3,3] $F.
% 1.80/2.06
% 1.80/2.06 Length of proof is 8. Level of proof is 5.
% 1.80/2.06
% 1.80/2.06 ---------------- PROOF ----------------
% 1.80/2.06 % SZS status Unsatisfiable
% 1.80/2.06 % SZS output start Refutation
% See solution above
% 1.80/2.06 ------------ end of proof -------------
% 1.80/2.06
% 1.80/2.06
% 1.80/2.06 Search stopped by max_proofs option.
% 1.80/2.06
% 1.80/2.06
% 1.80/2.06 Search stopped by max_proofs option.
% 1.80/2.06
% 1.80/2.06 ============ end of search ============
% 1.80/2.06
% 1.80/2.06 -------------- statistics -------------
% 1.80/2.06 clauses given 33
% 1.80/2.06 clauses generated 1188
% 1.80/2.06 clauses kept 359
% 1.80/2.06 clauses forward subsumed 699
% 1.80/2.06 clauses back subsumed 46
% 1.80/2.06 Kbytes malloced 1953
% 1.80/2.06
% 1.80/2.06 ----------- times (seconds) -----------
% 1.80/2.06 user CPU time 0.04 (0 hr, 0 min, 0 sec)
% 1.80/2.06 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.80/2.06 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.80/2.06
% 1.80/2.06 That finishes the proof of the theorem.
% 1.80/2.06
% 1.80/2.06 Process 32288 finished Wed Jul 27 09:19:49 2022
% 1.80/2.06 Otter interrupted
% 1.80/2.06 PROOF FOUND
%------------------------------------------------------------------------------