%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : COL055-1 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n011.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 12:48:06 EDT 2022
% Result : Unsatisfiable 1.81s 2.03s
% Output : Refutation 1.81s
% Verified :
% SZS Type : Refutation
% Derivation depth : 1
% Number of leaves : 2
% Syntax : Number of clauses : 3 ( 2 unt; 0 nHn; 3 RR)
% Number of literals : 4 ( 3 equ; 2 neg)
% Maximal clause size : 2 ( 1 avg)
% Maximal term depth : 2 ( 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 : 2 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( response(a,A) != B
| response(a,B) != A ),
file('COL055-1.p',unknown),
[] ).
cnf(3,axiom,
response(a,b) = b,
file('COL055-1.p',unknown),
[] ).
cnf(5,plain,
$false,
inference(hyper,[status(thm)],[3,1,3]),
[iquote('hyper,3,1,3')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : COL055-1 : TPTP v8.1.0. Released v1.0.0.
% 0.00/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n011.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 02:18:10 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.81/2.03 ----- Otter 3.3f, August 2004 -----
% 1.81/2.03 The process was started by sandbox on n011.cluster.edu,
% 1.81/2.03 Wed Jul 27 02:18:10 2022
% 1.81/2.03 The command was "./otter". The process ID is 3055.
% 1.81/2.03
% 1.81/2.03 set(prolog_style_variables).
% 1.81/2.03 set(auto).
% 1.81/2.03 dependent: set(auto1).
% 1.81/2.03 dependent: set(process_input).
% 1.81/2.03 dependent: clear(print_kept).
% 1.81/2.03 dependent: clear(print_new_demod).
% 1.81/2.03 dependent: clear(print_back_demod).
% 1.81/2.03 dependent: clear(print_back_sub).
% 1.81/2.03 dependent: set(control_memory).
% 1.81/2.03 dependent: assign(max_mem, 12000).
% 1.81/2.03 dependent: assign(pick_given_ratio, 4).
% 1.81/2.03 dependent: assign(stats_level, 1).
% 1.81/2.03 dependent: assign(max_seconds, 10800).
% 1.81/2.03 clear(print_given).
% 1.81/2.03
% 1.81/2.03 list(usable).
% 1.81/2.03 0 [] A=A.
% 1.81/2.03 0 [] response(a,b)=b.
% 1.81/2.03 0 [] response(a,Z)!=W|response(a,W)!=Z.
% 1.81/2.03 end_of_list.
% 1.81/2.03
% 1.81/2.03 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=2.
% 1.81/2.03
% 1.81/2.03 This is a Horn set with equality. The strategy will be
% 1.81/2.03 Knuth-Bendix and hyper_res, with positive clauses in
% 1.81/2.03 sos and nonpositive clauses in usable.
% 1.81/2.03
% 1.81/2.03 dependent: set(knuth_bendix).
% 1.81/2.03 dependent: set(anl_eq).
% 1.81/2.03 dependent: set(para_from).
% 1.81/2.03 dependent: set(para_into).
% 1.81/2.03 dependent: clear(para_from_right).
% 1.81/2.03 dependent: clear(para_into_right).
% 1.81/2.03 dependent: set(para_from_vars).
% 1.81/2.03 dependent: set(eq_units_both_ways).
% 1.81/2.03 dependent: set(dynamic_demod_all).
% 1.81/2.03 dependent: set(dynamic_demod).
% 1.81/2.03 dependent: set(order_eq).
% 1.81/2.03 dependent: set(back_demod).
% 1.81/2.03 dependent: set(lrpo).
% 1.81/2.03 dependent: set(hyper_res).
% 1.81/2.03 dependent: clear(order_hyper).
% 1.81/2.03
% 1.81/2.03 ------------> process usable:
% 1.81/2.03 ** KEPT (pick-wt=10): 1 [] response(a,A)!=B|response(a,B)!=A.
% 1.81/2.03
% 1.81/2.03 ------------> process sos:
% 1.81/2.03 ** KEPT (pick-wt=3): 2 [] A=A.
% 1.81/2.03 ** KEPT (pick-wt=5): 3 [] response(a,b)=b.
% 1.81/2.03 ---> New Demodulator: 4 [new_demod,3] response(a,b)=b.
% 1.81/2.03 Following clause subsumed by 2 during input processing: 0 [copy,2,flip.1] A=A.
% 1.81/2.03 >>>> Starting back demodulation with 4.
% 1.81/2.03
% 1.81/2.03 ======= end of input processing =======
% 1.81/2.03
% 1.81/2.03 =========== start of search ===========
% 1.81/2.03
% 1.81/2.03 -------- PROOF --------
% 1.81/2.03
% 1.81/2.03 -----> EMPTY CLAUSE at 0.00 sec ----> 5 [hyper,3,1,3] $F.
% 1.81/2.03
% 1.81/2.03 Length of proof is 0. Level of proof is 0.
% 1.81/2.03
% 1.81/2.03 ---------------- PROOF ----------------
% 1.81/2.03 % SZS status Unsatisfiable
% 1.81/2.03 % SZS output start Refutation
% See solution above
% 1.81/2.03 ------------ end of proof -------------
% 1.81/2.03
% 1.81/2.03
% 1.81/2.03 Search stopped by max_proofs option.
% 1.81/2.03
% 1.81/2.03
% 1.81/2.03 Search stopped by max_proofs option.
% 1.81/2.03
% 1.81/2.03 ============ end of search ============
% 1.81/2.03
% 1.81/2.03 -------------- statistics -------------
% 1.81/2.03 clauses given 2
% 1.81/2.03 clauses generated 1
% 1.81/2.03 clauses kept 3
% 1.81/2.03 clauses forward subsumed 1
% 1.81/2.03 clauses back subsumed 0
% 1.81/2.03 Kbytes malloced 976
% 1.81/2.03
% 1.81/2.03 ----------- times (seconds) -----------
% 1.81/2.03 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.81/2.03 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.81/2.03 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.81/2.03
% 1.81/2.03 That finishes the proof of the theorem.
% 1.81/2.03
% 1.81/2.03 Process 3055 finished Wed Jul 27 02:18:12 2022
% 1.81/2.03 Otter interrupted
% 1.81/2.03 PROOF FOUND
%------------------------------------------------------------------------------