TSTP Solution File: HEN005-5 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : HEN005-5 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n009.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 12:57:56 EDT 2022
% Result : Unsatisfiable 1.85s 1.99s
% Output : Refutation 1.85s
% Verified :
% SZS Type : Refutation
% Derivation depth : 5
% Number of leaves : 7
% Syntax : Number of clauses : 16 ( 11 unt; 0 nHn; 11 RR)
% Number of literals : 24 ( 23 equ; 12 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 4 ( 1 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 5 ( 5 usr; 4 con; 0-2 aty)
% Number of variables : 11 ( 1 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( divide(A,B) != zero
| divide(B,A) != zero
| A = B ),
file('HEN005-5.p',unknown),
[] ).
cnf(2,axiom,
divide(a,c) != zero,
file('HEN005-5.p',unknown),
[] ).
cnf(3,axiom,
A = A,
file('HEN005-5.p',unknown),
[] ).
cnf(6,axiom,
divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B)) = zero,
file('HEN005-5.p',unknown),
[] ).
cnf(9,axiom,
divide(zero,A) = zero,
file('HEN005-5.p',unknown),
[] ).
cnf(12,axiom,
divide(a,b) = zero,
file('HEN005-5.p',unknown),
[] ).
cnf(14,axiom,
divide(b,c) = zero,
file('HEN005-5.p',unknown),
[] ).
cnf(17,plain,
( A = zero
| zero != zero
| divide(A,zero) != zero ),
inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[9,1]),9,9,9]),
[iquote('para_into,8.1.1,1.3.1,demod,9,9,9')] ).
cnf(18,plain,
( zero != zero
| divide(A,zero) != zero
| zero = A ),
inference(para_from,[status(thm),theory(equality)],[9,1]),
[iquote('para_from,8.1.1,1.1.1')] ).
cnf(34,plain,
( zero != zero
| divide(divide(a,c),zero) != zero ),
inference(para_from,[status(thm),theory(equality)],[17,2]),
[iquote('para_from,17.1.1,2.1.1')] ).
cnf(48,plain,
( zero != zero
| divide(divide(divide(a,c),zero),zero) != zero ),
inference(para_into,[status(thm),theory(equality)],[34,17]),
[iquote('para_into,34.2.1,17.1.1')] ).
cnf(73,plain,
divide(divide(divide(a,A),divide(b,A)),zero) = zero,
inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[6,12]),9]),
[iquote('para_into,6.1.1.2.1,12.1.1,demod,9')] ).
cnf(348,plain,
divide(divide(a,A),divide(b,A)) = zero,
inference(flip,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[73,18,3])]),
[iquote('hyper,73,18,3,flip.1')] ).
cnf(372,plain,
divide(divide(a,c),zero) = zero,
inference(para_into,[status(thm),theory(equality)],[348,14]),
[iquote('para_into,348.1.1.2,14.1.1')] ).
cnf(375,plain,
zero != zero,
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[48]),372,9]),
[iquote('back_demod,48,demod,372,9')] ).
cnf(376,plain,
$false,
inference(binary,[status(thm)],[375,3]),
[iquote('binary,375.1,3.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.13 % Problem : HEN005-5 : TPTP v8.1.0. Released v1.0.0.
% 0.12/0.13 % Command : otter-tptp-script %s
% 0.13/0.35 % Computer : n009.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Wed Jul 27 08:54:51 EDT 2022
% 0.13/0.35 % CPUTime :
% 1.85/1.99 ----- Otter 3.3f, August 2004 -----
% 1.85/1.99 The process was started by sandbox on n009.cluster.edu,
% 1.85/1.99 Wed Jul 27 08:54:51 2022
% 1.85/1.99 The command was "./otter". The process ID is 31142.
% 1.85/1.99
% 1.85/1.99 set(prolog_style_variables).
% 1.85/1.99 set(auto).
% 1.85/1.99 dependent: set(auto1).
% 1.85/1.99 dependent: set(process_input).
% 1.85/1.99 dependent: clear(print_kept).
% 1.85/1.99 dependent: clear(print_new_demod).
% 1.85/1.99 dependent: clear(print_back_demod).
% 1.85/1.99 dependent: clear(print_back_sub).
% 1.85/1.99 dependent: set(control_memory).
% 1.85/1.99 dependent: assign(max_mem, 12000).
% 1.85/1.99 dependent: assign(pick_given_ratio, 4).
% 1.85/1.99 dependent: assign(stats_level, 1).
% 1.85/1.99 dependent: assign(max_seconds, 10800).
% 1.85/1.99 clear(print_given).
% 1.85/1.99
% 1.85/1.99 list(usable).
% 1.85/1.99 0 [] A=A.
% 1.85/1.99 0 [] divide(divide(X,Y),X)=zero.
% 1.85/1.99 0 [] divide(divide(divide(X,Z),divide(Y,Z)),divide(divide(X,Y),Z))=zero.
% 1.85/1.99 0 [] divide(zero,X)=zero.
% 1.85/1.99 0 [] divide(X,Y)!=zero|divide(Y,X)!=zero|X=Y.
% 1.85/1.99 0 [] divide(X,identity)=zero.
% 1.85/1.99 0 [] divide(a,b)=zero.
% 1.85/1.99 0 [] divide(b,c)=zero.
% 1.85/1.99 0 [] divide(a,c)!=zero.
% 1.85/1.99 end_of_list.
% 1.85/1.99
% 1.85/1.99 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=3.
% 1.85/1.99
% 1.85/1.99 This is a Horn set with equality. The strategy will be
% 1.85/1.99 Knuth-Bendix and hyper_res, with positive clauses in
% 1.85/1.99 sos and nonpositive clauses in usable.
% 1.85/1.99
% 1.85/1.99 dependent: set(knuth_bendix).
% 1.85/1.99 dependent: set(anl_eq).
% 1.85/1.99 dependent: set(para_from).
% 1.85/1.99 dependent: set(para_into).
% 1.85/1.99 dependent: clear(para_from_right).
% 1.85/1.99 dependent: clear(para_into_right).
% 1.85/1.99 dependent: set(para_from_vars).
% 1.85/1.99 dependent: set(eq_units_both_ways).
% 1.85/1.99 dependent: set(dynamic_demod_all).
% 1.85/1.99 dependent: set(dynamic_demod).
% 1.85/1.99 dependent: set(order_eq).
% 1.85/1.99 dependent: set(back_demod).
% 1.85/1.99 dependent: set(lrpo).
% 1.85/1.99 dependent: set(hyper_res).
% 1.85/1.99 dependent: clear(order_hyper).
% 1.85/1.99
% 1.85/1.99 ------------> process usable:
% 1.85/1.99 ** KEPT (pick-wt=13): 1 [] divide(A,B)!=zero|divide(B,A)!=zero|A=B.
% 1.85/1.99 ** KEPT (pick-wt=5): 2 [] divide(a,c)!=zero.
% 1.85/1.99
% 1.85/1.99 ------------> process sos:
% 1.85/1.99 ** KEPT (pick-wt=3): 3 [] A=A.
% 1.85/1.99 ** KEPT (pick-wt=7): 4 [] divide(divide(A,B),A)=zero.
% 1.85/1.99 ---> New Demodulator: 5 [new_demod,4] divide(divide(A,B),A)=zero.
% 1.85/1.99 ** KEPT (pick-wt=15): 6 [] divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B))=zero.
% 1.85/1.99 ---> New Demodulator: 7 [new_demod,6] divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B))=zero.
% 1.85/1.99 ** KEPT (pick-wt=5): 8 [] divide(zero,A)=zero.
% 1.85/1.99 ---> New Demodulator: 9 [new_demod,8] divide(zero,A)=zero.
% 1.85/1.99 ** KEPT (pick-wt=5): 10 [] divide(A,identity)=zero.
% 1.85/1.99 ---> New Demodulator: 11 [new_demod,10] divide(A,identity)=zero.
% 1.85/1.99 ** KEPT (pick-wt=5): 12 [] divide(a,b)=zero.
% 1.85/1.99 ---> New Demodulator: 13 [new_demod,12] divide(a,b)=zero.
% 1.85/1.99 ** KEPT (pick-wt=5): 14 [] divide(b,c)=zero.
% 1.85/1.99 ---> New Demodulator: 15 [new_demod,14] divide(b,c)=zero.
% 1.85/1.99 Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.85/1.99 >>>> Starting back demodulation with 5.
% 1.85/1.99 >>>> Starting back demodulation with 7.
% 1.85/1.99 >>>> Starting back demodulation with 9.
% 1.85/1.99 >>>> Starting back demodulation with 11.
% 1.85/1.99 >>>> Starting back demodulation with 13.
% 1.85/1.99 >>>> Starting back demodulation with 15.
% 1.85/1.99
% 1.85/1.99 ======= end of input processing =======
% 1.85/1.99
% 1.85/1.99 =========== start of search ===========
% 1.85/1.99
% 1.85/1.99 -------- PROOF --------
% 1.85/1.99
% 1.85/1.99 ----> UNIT CONFLICT at 0.05 sec ----> 376 [binary,375.1,3.1] $F.
% 1.85/1.99
% 1.85/1.99 Length of proof is 8. Level of proof is 4.
% 1.85/1.99
% 1.85/1.99 ---------------- PROOF ----------------
% 1.85/1.99 % SZS status Unsatisfiable
% 1.85/1.99 % SZS output start Refutation
% See solution above
% 1.85/1.99 ------------ end of proof -------------
% 1.85/1.99
% 1.85/1.99
% 1.85/1.99 Search stopped by max_proofs option.
% 1.85/1.99
% 1.85/1.99
% 1.85/1.99 Search stopped by max_proofs option.
% 1.85/1.99
% 1.85/1.99 ============ end of search ============
% 1.85/1.99
% 1.85/1.99 -------------- statistics -------------
% 1.85/1.99 clauses given 42
% 1.85/1.99 clauses generated 968
% 1.85/1.99 clauses kept 351
% 1.85/1.99 clauses forward subsumed 497
% 1.85/1.99 clauses back subsumed 71
% 1.85/1.99 Kbytes malloced 1953
% 1.85/1.99
% 1.85/1.99 ----------- times (seconds) -----------
% 1.85/1.99 user CPU time 0.05 (0 hr, 0 min, 0 sec)
% 1.85/1.99 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.85/1.99 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.85/1.99
% 1.85/1.99 That finishes the proof of the theorem.
% 1.85/1.99
% 1.85/1.99 Process 31142 finished Wed Jul 27 08:54:53 2022
% 1.85/1.99 Otter interrupted
% 1.85/1.99 PROOF FOUND
%------------------------------------------------------------------------------