TSTP Solution File: HEN003-5 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : HEN003-5 : TPTP v8.1.0. Released v1.0.0.
% 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 12:57:54 EDT 2022
% Result : Unsatisfiable 1.72s 1.96s
% Output : Refutation 1.72s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 6
% Syntax : Number of clauses : 11 ( 9 unt; 0 nHn; 4 RR)
% Number of literals : 15 ( 14 equ; 5 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 : 3 ( 3 usr; 2 con; 0-2 aty)
% Number of variables : 15 ( 4 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( divide(A,B) != zero
| divide(B,A) != zero
| A = B ),
file('HEN003-5.p',unknown),
[] ).
cnf(2,axiom,
divide(a,a) != zero,
file('HEN003-5.p',unknown),
[] ).
cnf(3,axiom,
A = A,
file('HEN003-5.p',unknown),
[] ).
cnf(5,axiom,
divide(divide(A,B),A) = zero,
file('HEN003-5.p',unknown),
[] ).
cnf(6,axiom,
divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B)) = zero,
file('HEN003-5.p',unknown),
[] ).
cnf(8,axiom,
divide(zero,A) = zero,
file('HEN003-5.p',unknown),
[] ).
cnf(14,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(48,plain,
divide(divide(divide(A,A),divide(B,A)),zero) = zero,
inference(para_into,[status(thm),theory(equality)],[6,5]),
[iquote('para_into,6.1.1.2,4.1.1')] ).
cnf(208,plain,
divide(divide(A,A),divide(B,A)) = zero,
inference(flip,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[48,14,3])]),
[iquote('hyper,48,14,3,flip.1')] ).
cnf(210,plain,
divide(A,A) = zero,
inference(flip,[status(thm),theory(equality)],[inference(demod,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[208,1,5]),5])]),
[iquote('hyper,208,1,4,demod,5,flip.1')] ).
cnf(212,plain,
$false,
inference(binary,[status(thm)],[210,2]),
[iquote('binary,210.1,2.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12 % Problem : HEN003-5 : TPTP v8.1.0. Released v1.0.0.
% 0.06/0.12 % 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 09:02:16 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.72/1.96 ----- Otter 3.3f, August 2004 -----
% 1.72/1.96 The process was started by sandbox2 on n018.cluster.edu,
% 1.72/1.96 Wed Jul 27 09:02:16 2022
% 1.72/1.96 The command was "./otter". The process ID is 30925.
% 1.72/1.96
% 1.72/1.96 set(prolog_style_variables).
% 1.72/1.96 set(auto).
% 1.72/1.96 dependent: set(auto1).
% 1.72/1.96 dependent: set(process_input).
% 1.72/1.96 dependent: clear(print_kept).
% 1.72/1.96 dependent: clear(print_new_demod).
% 1.72/1.96 dependent: clear(print_back_demod).
% 1.72/1.96 dependent: clear(print_back_sub).
% 1.72/1.96 dependent: set(control_memory).
% 1.72/1.96 dependent: assign(max_mem, 12000).
% 1.72/1.96 dependent: assign(pick_given_ratio, 4).
% 1.72/1.96 dependent: assign(stats_level, 1).
% 1.72/1.96 dependent: assign(max_seconds, 10800).
% 1.72/1.96 clear(print_given).
% 1.72/1.96
% 1.72/1.96 list(usable).
% 1.72/1.96 0 [] A=A.
% 1.72/1.96 0 [] divide(divide(X,Y),X)=zero.
% 1.72/1.96 0 [] divide(divide(divide(X,Z),divide(Y,Z)),divide(divide(X,Y),Z))=zero.
% 1.72/1.96 0 [] divide(zero,X)=zero.
% 1.72/1.96 0 [] divide(X,Y)!=zero|divide(Y,X)!=zero|X=Y.
% 1.72/1.96 0 [] divide(X,identity)=zero.
% 1.72/1.96 0 [] divide(a,a)!=zero.
% 1.72/1.96 end_of_list.
% 1.72/1.96
% 1.72/1.96 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=3.
% 1.72/1.96
% 1.72/1.96 This is a Horn set with equality. The strategy will be
% 1.72/1.96 Knuth-Bendix and hyper_res, with positive clauses in
% 1.72/1.96 sos and nonpositive clauses in usable.
% 1.72/1.96
% 1.72/1.96 dependent: set(knuth_bendix).
% 1.72/1.96 dependent: set(anl_eq).
% 1.72/1.96 dependent: set(para_from).
% 1.72/1.96 dependent: set(para_into).
% 1.72/1.96 dependent: clear(para_from_right).
% 1.72/1.96 dependent: clear(para_into_right).
% 1.72/1.96 dependent: set(para_from_vars).
% 1.72/1.96 dependent: set(eq_units_both_ways).
% 1.72/1.96 dependent: set(dynamic_demod_all).
% 1.72/1.96 dependent: set(dynamic_demod).
% 1.72/1.96 dependent: set(order_eq).
% 1.72/1.96 dependent: set(back_demod).
% 1.72/1.96 dependent: set(lrpo).
% 1.72/1.96 dependent: set(hyper_res).
% 1.72/1.96 dependent: clear(order_hyper).
% 1.72/1.96
% 1.72/1.96 ------------> process usable:
% 1.72/1.96 ** KEPT (pick-wt=13): 1 [] divide(A,B)!=zero|divide(B,A)!=zero|A=B.
% 1.72/1.96 ** KEPT (pick-wt=5): 2 [] divide(a,a)!=zero.
% 1.72/1.96
% 1.72/1.96 ------------> process sos:
% 1.72/1.96 ** KEPT (pick-wt=3): 3 [] A=A.
% 1.72/1.96 ** KEPT (pick-wt=7): 4 [] divide(divide(A,B),A)=zero.
% 1.72/1.96 ---> New Demodulator: 5 [new_demod,4] divide(divide(A,B),A)=zero.
% 1.72/1.96 ** KEPT (pick-wt=15): 6 [] divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B))=zero.
% 1.72/1.96 ---> New Demodulator: 7 [new_demod,6] divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B))=zero.
% 1.72/1.96 ** KEPT (pick-wt=5): 8 [] divide(zero,A)=zero.
% 1.72/1.96 ---> New Demodulator: 9 [new_demod,8] divide(zero,A)=zero.
% 1.72/1.96 ** KEPT (pick-wt=5): 10 [] divide(A,identity)=zero.
% 1.72/1.96 ---> New Demodulator: 11 [new_demod,10] divide(A,identity)=zero.
% 1.72/1.96 Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.72/1.96 >>>> Starting back demodulation with 5.
% 1.72/1.96 >>>> Starting back demodulation with 7.
% 1.72/1.96 >>>> Starting back demodulation with 9.
% 1.72/1.96 >>>> Starting back demodulation with 11.
% 1.72/1.96
% 1.72/1.96 ======= end of input processing =======
% 1.72/1.96
% 1.72/1.96 =========== start of search ===========
% 1.72/1.96
% 1.72/1.96 -------- PROOF --------
% 1.72/1.96
% 1.72/1.96 ----> UNIT CONFLICT at 0.02 sec ----> 212 [binary,210.1,2.1] $F.
% 1.72/1.96
% 1.72/1.96 Length of proof is 4. Level of proof is 3.
% 1.72/1.96
% 1.72/1.96 ---------------- PROOF ----------------
% 1.72/1.96 % SZS status Unsatisfiable
% 1.72/1.96 % SZS output start Refutation
% See solution above
% 1.72/1.96 ------------ end of proof -------------
% 1.72/1.96
% 1.72/1.96
% 1.72/1.96 Search stopped by max_proofs option.
% 1.72/1.96
% 1.72/1.96
% 1.72/1.96 Search stopped by max_proofs option.
% 1.72/1.96
% 1.72/1.96 ============ end of search ============
% 1.72/1.96
% 1.72/1.96 -------------- statistics -------------
% 1.72/1.96 clauses given 25
% 1.72/1.96 clauses generated 552
% 1.72/1.96 clauses kept 193
% 1.72/1.96 clauses forward subsumed 300
% 1.72/1.96 clauses back subsumed 18
% 1.72/1.96 Kbytes malloced 976
% 1.72/1.96
% 1.72/1.96 ----------- times (seconds) -----------
% 1.72/1.96 user CPU time 0.02 (0 hr, 0 min, 0 sec)
% 1.72/1.96 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.72/1.96 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.72/1.96
% 1.72/1.96 That finishes the proof of the theorem.
% 1.72/1.96
% 1.72/1.96 Process 30925 finished Wed Jul 27 09:02:18 2022
% 1.72/1.96 Otter interrupted
% 1.72/1.96 PROOF FOUND
%------------------------------------------------------------------------------