%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : LDA007-3 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n017.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 13:05:49 EDT 2022
% Result : Unsatisfiable 1.69s 1.91s
% Output : Refutation 1.69s
% Verified :
% SZS Type : Refutation
% Derivation depth : 5
% Number of leaves : 7
% Syntax : Number of clauses : 19 ( 19 unt; 0 nHn; 15 RR)
% Number of literals : 19 ( 18 equ; 2 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 9 ( 9 usr; 8 con; 0-2 aty)
% Number of variables : 8 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
f(t,tsk) != f(tt_ts,tk),
file('LDA007-3.p',unknown),
[] ).
cnf(2,plain,
f(tt_ts,tk) != f(t,tsk),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[1])]),
[iquote('copy,1,flip.1')] ).
cnf(4,axiom,
f(A,f(B,C)) = f(f(A,B),f(A,C)),
file('LDA007-3.p',unknown),
[] ).
cnf(5,plain,
f(f(A,B),f(A,C)) = f(A,f(B,C)),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[4])]),
[iquote('copy,4,flip.1')] ).
cnf(7,axiom,
tt = f(t,t),
file('LDA007-3.p',unknown),
[] ).
cnf(8,plain,
f(t,t) = tt,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[7])]),
[iquote('copy,7,flip.1')] ).
cnf(10,axiom,
ts = f(t,s),
file('LDA007-3.p',unknown),
[] ).
cnf(12,plain,
f(t,s) = ts,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[10])]),
[iquote('copy,10,flip.1')] ).
cnf(13,axiom,
tt_ts = f(tt,ts),
file('LDA007-3.p',unknown),
[] ).
cnf(15,plain,
f(tt,ts) = tt_ts,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[13])]),
[iquote('copy,13,flip.1')] ).
cnf(16,axiom,
tk = f(t,k),
file('LDA007-3.p',unknown),
[] ).
cnf(17,plain,
f(t,k) = tk,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[16])]),
[iquote('copy,16,flip.1')] ).
cnf(19,axiom,
tsk = f(ts,k),
file('LDA007-3.p',unknown),
[] ).
cnf(21,plain,
f(ts,k) = tsk,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[19])]),
[iquote('copy,19,flip.1')] ).
cnf(28,plain,
f(tt,f(t,A)) = f(t,f(t,A)),
inference(para_into,[status(thm),theory(equality)],[5,8]),
[iquote('para_into,5.1.1.1,8.1.1')] ).
cnf(32,plain,
f(f(t,A),tk) = f(t,f(A,k)),
inference(para_into,[status(thm),theory(equality)],[5,17]),
[iquote('para_into,5.1.1.2,17.1.1')] ).
cnf(76,plain,
f(t,ts) = tt_ts,
inference(flip,[status(thm),theory(equality)],[inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[28,12]),15,12])]),
[iquote('para_into,28.1.1.2,11.1.1,demod,15,12,flip.1')] ).
cnf(200,plain,
f(tt_ts,tk) = f(t,tsk),
inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[32,76]),21]),
[iquote('para_into,32.1.1.1,76.1.1,demod,21')] ).
cnf(202,plain,
$false,
inference(binary,[status(thm)],[200,2]),
[iquote('binary,200.1,2.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.11 % Problem : LDA007-3 : TPTP v8.1.0. Released v1.0.0.
% 0.06/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n017.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 01:31:18 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.69/1.91 ----- Otter 3.3f, August 2004 -----
% 1.69/1.91 The process was started by sandbox on n017.cluster.edu,
% 1.69/1.91 Wed Jul 27 01:31:18 2022
% 1.69/1.91 The command was "./otter". The process ID is 5565.
% 1.69/1.91
% 1.69/1.91 set(prolog_style_variables).
% 1.69/1.91 set(auto).
% 1.69/1.91 dependent: set(auto1).
% 1.69/1.91 dependent: set(process_input).
% 1.69/1.91 dependent: clear(print_kept).
% 1.69/1.91 dependent: clear(print_new_demod).
% 1.69/1.91 dependent: clear(print_back_demod).
% 1.69/1.91 dependent: clear(print_back_sub).
% 1.69/1.91 dependent: set(control_memory).
% 1.69/1.91 dependent: assign(max_mem, 12000).
% 1.69/1.91 dependent: assign(pick_given_ratio, 4).
% 1.69/1.91 dependent: assign(stats_level, 1).
% 1.69/1.91 dependent: assign(max_seconds, 10800).
% 1.69/1.91 clear(print_given).
% 1.69/1.91
% 1.69/1.91 list(usable).
% 1.69/1.91 0 [] A=A.
% 1.69/1.91 0 [] f(X,f(Y,Z))=f(f(X,Y),f(X,Z)).
% 1.69/1.91 0 [] tt=f(t,t).
% 1.69/1.91 0 [] ts=f(t,s).
% 1.69/1.91 0 [] tt_ts=f(tt,ts).
% 1.69/1.91 0 [] tk=f(t,k).
% 1.69/1.91 0 [] tsk=f(ts,k).
% 1.69/1.91 0 [] f(t,tsk)!=f(tt_ts,tk).
% 1.69/1.91 end_of_list.
% 1.69/1.91
% 1.69/1.91 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.69/1.91
% 1.69/1.91 All clauses are units, and equality is present; the
% 1.69/1.91 strategy will be Knuth-Bendix with positive clauses in sos.
% 1.69/1.91
% 1.69/1.91 dependent: set(knuth_bendix).
% 1.69/1.91 dependent: set(anl_eq).
% 1.69/1.91 dependent: set(para_from).
% 1.69/1.91 dependent: set(para_into).
% 1.69/1.91 dependent: clear(para_from_right).
% 1.69/1.91 dependent: clear(para_into_right).
% 1.69/1.91 dependent: set(para_from_vars).
% 1.69/1.91 dependent: set(eq_units_both_ways).
% 1.69/1.91 dependent: set(dynamic_demod_all).
% 1.69/1.91 dependent: set(dynamic_demod).
% 1.69/1.91 dependent: set(order_eq).
% 1.69/1.91 dependent: set(back_demod).
% 1.69/1.91 dependent: set(lrpo).
% 1.69/1.91
% 1.69/1.91 ------------> process usable:
% 1.69/1.91 ** KEPT (pick-wt=7): 2 [copy,1,flip.1] f(tt_ts,tk)!=f(t,tsk).
% 1.69/1.91
% 1.69/1.91 ------------> process sos:
% 1.69/1.91 ** KEPT (pick-wt=3): 3 [] A=A.
% 1.69/1.91 ** KEPT (pick-wt=13): 5 [copy,4,flip.1] f(f(A,B),f(A,C))=f(A,f(B,C)).
% 1.69/1.91 ---> New Demodulator: 6 [new_demod,5] f(f(A,B),f(A,C))=f(A,f(B,C)).
% 1.69/1.91 ** KEPT (pick-wt=5): 8 [copy,7,flip.1] f(t,t)=tt.
% 1.69/1.91 ---> New Demodulator: 9 [new_demod,8] f(t,t)=tt.
% 1.69/1.91 ** KEPT (pick-wt=5): 11 [copy,10,flip.1] f(t,s)=ts.
% 1.69/1.91 ---> New Demodulator: 12 [new_demod,11] f(t,s)=ts.
% 1.69/1.91 ** KEPT (pick-wt=5): 14 [copy,13,flip.1] f(tt,ts)=tt_ts.
% 1.69/1.91 ---> New Demodulator: 15 [new_demod,14] f(tt,ts)=tt_ts.
% 1.69/1.91 ** KEPT (pick-wt=5): 17 [copy,16,flip.1] f(t,k)=tk.
% 1.69/1.91 ---> New Demodulator: 18 [new_demod,17] f(t,k)=tk.
% 1.69/1.91 ** KEPT (pick-wt=5): 20 [copy,19,flip.1] f(ts,k)=tsk.
% 1.69/1.91 ---> New Demodulator: 21 [new_demod,20] f(ts,k)=tsk.
% 1.69/1.91 Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.69/1.91 >>>> Starting back demodulation with 6.
% 1.69/1.91 >>>> Starting back demodulation with 9.
% 1.69/1.91 >>>> Starting back demodulation with 12.
% 1.69/1.91 >>>> Starting back demodulation with 15.
% 1.69/1.91 >>>> Starting back demodulation with 18.
% 1.69/1.91 >>>> Starting back demodulation with 21.
% 1.69/1.91
% 1.69/1.91 ======= end of input processing =======
% 1.69/1.91
% 1.69/1.91 =========== start of search ===========
% 1.69/1.91
% 1.69/1.91 -------- PROOF --------
% 1.69/1.91
% 1.69/1.91 ----> UNIT CONFLICT at 0.00 sec ----> 202 [binary,200.1,2.1] $F.
% 1.69/1.91
% 1.69/1.91 Length of proof is 11. Level of proof is 4.
% 1.69/1.91
% 1.69/1.91 ---------------- PROOF ----------------
% 1.69/1.91 % SZS status Unsatisfiable
% 1.69/1.91 % SZS output start Refutation
% See solution above
% 1.69/1.91 ------------ end of proof -------------
% 1.69/1.91
% 1.69/1.91
% 1.69/1.91 Search stopped by max_proofs option.
% 1.69/1.91
% 1.69/1.91
% 1.69/1.91 Search stopped by max_proofs option.
% 1.69/1.91
% 1.69/1.91 ============ end of search ============
% 1.69/1.91
% 1.69/1.91 -------------- statistics -------------
% 1.69/1.91 clauses given 26
% 1.69/1.91 clauses generated 174
% 1.69/1.91 clauses kept 103
% 1.69/1.91 clauses forward subsumed 96
% 1.69/1.91 clauses back subsumed 0
% 1.69/1.91 Kbytes malloced 1953
% 1.69/1.91
% 1.69/1.91 ----------- times (seconds) -----------
% 1.69/1.91 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.69/1.91 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.69/1.91 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.69/1.91
% 1.69/1.91 That finishes the proof of the theorem.
% 1.69/1.91
% 1.69/1.91 Process 5565 finished Wed Jul 27 01:31:19 2022
% 1.69/1.91 Otter interrupted
% 1.69/1.91 PROOF FOUND
%------------------------------------------------------------------------------