%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SWX186+1 : TPTP v9.3.0. Released v9.3.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n002.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 : Tue May 5 07:05:22 PM UTC 2026
% Result : Theorem 1.89s 2.11s
% Output : Refutation 1.89s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 7
% Syntax : Number of clauses : 12 ( 8 unt; 0 nHn; 7 RR)
% Number of literals : 16 ( 15 equ; 7 neg)
% Maximal clause size : 2 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 6 ( 6 usr; 2 con; 0-2 aty)
% Number of variables : 21 ( 9 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
nil != cons(A,B),
file('SWX186+1.p',unknown),
[] ).
cnf(2,plain,
cons(A,B) != nil,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[1])]),
[iquote('copy,1,flip.1')] ).
cnf(4,axiom,
( drop(A,B) != drop(A,C)
| B = C ),
file('SWX186+1.p',unknown),
[] ).
cnf(5,axiom,
A = A,
file('SWX186+1.p',unknown),
[] ).
cnf(11,axiom,
proj1S(s(A)) = A,
file('SWX186+1.p',unknown),
[] ).
cnf(12,axiom,
drop(s(A),nil) = nil,
file('SWX186+1.p',unknown),
[] ).
cnf(15,axiom,
drop(s(A),cons(B,C)) = drop(A,C),
file('SWX186+1.p',unknown),
[] ).
cnf(16,axiom,
drop(z,A) = A,
file('SWX186+1.p',unknown),
[] ).
cnf(19,plain,
( A = B
| drop(C,B) != drop(C,A) ),
inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[11,4]),11]),
[iquote('para_into,10.1.1,4.2.1,demod,11')] ).
cnf(38,plain,
( A = nil
| drop(s(B),A) != nil ),
inference(flip,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[19,12])]),
[iquote('para_into,19.2.1,12.1.1,flip.2')] ).
cnf(254,plain,
( nil != nil
| drop(A,B) != nil ),
inference(demod,[status(thm),theory(equality)],[inference(para_from,[status(thm),theory(equality)],[38,2]),15]),
[iquote('para_from,38.1.1,2.1.1,demod,15')] ).
cnf(255,plain,
$false,
inference(hyper,[status(thm)],[254,5,16]),
[iquote('hyper,254,5,16')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.12 % Problem : SWX186+1 : TPTP v9.3.0. Released v9.3.0.
% 0.12/0.13 % Command : otter-tptp-script %s
% 0.17/0.34 % Computer : n002.cluster.edu
% 0.17/0.34 % Model : x86_64 x86_64
% 0.17/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.17/0.34 % Memory : 8042.1875MB
% 0.17/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.17/0.34 % CPULimit : 300
% 0.17/0.34 % WCLimit : 300
% 0.17/0.34 % DateTime : Tue May 5 09:37:31 EDT 2026
% 0.17/0.35 % CPUTime :
% 1.89/2.11 ----- Otter 3.3f, August 2004 -----
% 1.89/2.11 The process was started by sandbox on n002.cluster.edu,
% 1.89/2.11 Tue May 5 09:37:31 2026
% 1.89/2.11 The command was "./otter". The process ID is 22345.
% 1.89/2.11
% 1.89/2.11 set(prolog_style_variables).
% 1.89/2.11 set(auto).
% 1.89/2.11 dependent: set(auto1).
% 1.89/2.11 dependent: set(process_input).
% 1.89/2.11 dependent: clear(print_kept).
% 1.89/2.11 dependent: clear(print_new_demod).
% 1.89/2.11 dependent: clear(print_back_demod).
% 1.89/2.11 dependent: clear(print_back_sub).
% 1.89/2.11 dependent: set(control_memory).
% 1.89/2.11 dependent: assign(max_mem, 12000).
% 1.89/2.11 dependent: assign(pick_given_ratio, 4).
% 1.89/2.11 dependent: assign(stats_level, 1).
% 1.89/2.11 dependent: assign(max_seconds, 10800).
% 1.89/2.11 clear(print_given).
% 1.89/2.11
% 1.89/2.11 formula_list(usable).
% 1.89/2.11 all A (A=A).
% 1.89/2.11 all X X2 (head(cons(X,X2))=X).
% 1.89/2.11 all X X2 (tail(cons(X,X2))=X2).
% 1.89/2.11 all X X2 (nil!=cons(X,X2)).
% 1.89/2.11 all X (proj1S(s(X))=X).
% 1.89/2.11 all X (s(X)!=z).
% 1.89/2.11 all Z (drop(s(Z),nil)=nil).
% 1.89/2.11 all Z X2 X3 (drop(s(Z),cons(X2,X3))=drop(Z,X3)).
% 1.89/2.11 all Y (drop(z,Y)=Y).
% 1.89/2.11 -(exists N Xs Ys (-(drop(N,Xs)=drop(N,Ys)->Xs=Ys))).
% 1.89/2.11 end_of_list.
% 1.89/2.11
% 1.89/2.11 -------> usable clausifies to:
% 1.89/2.11
% 1.89/2.11 list(usable).
% 1.89/2.11 0 [] A=A.
% 1.89/2.11 0 [] head(cons(X,X2))=X.
% 1.89/2.11 0 [] tail(cons(X,X2))=X2.
% 1.89/2.11 0 [] nil!=cons(X,X2).
% 1.89/2.11 0 [] proj1S(s(X))=X.
% 1.89/2.11 0 [] s(X)!=z.
% 1.89/2.11 0 [] drop(s(Z),nil)=nil.
% 1.89/2.11 0 [] drop(s(Z),cons(X2,X3))=drop(Z,X3).
% 1.89/2.11 0 [] drop(z,Y)=Y.
% 1.89/2.11 0 [] drop(N,Xs)!=drop(N,Ys)|Xs=Ys.
% 1.89/2.11 end_of_list.
% 1.89/2.11
% 1.89/2.11 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=2.
% 1.89/2.11
% 1.89/2.11 This is a Horn set with equality. The strategy will be
% 1.89/2.11 Knuth-Bendix and hyper_res, with positive clauses in
% 1.89/2.11 sos and nonpositive clauses in usable.
% 1.89/2.11
% 1.89/2.11 dependent: set(knuth_bendix).
% 1.89/2.11 dependent: set(anl_eq).
% 1.89/2.11 dependent: set(para_from).
% 1.89/2.11 dependent: set(para_into).
% 1.89/2.11 dependent: clear(para_from_right).
% 1.89/2.11 dependent: clear(para_into_right).
% 1.89/2.11 dependent: set(para_from_vars).
% 1.89/2.11 dependent: set(eq_units_both_ways).
% 1.89/2.11 dependent: set(dynamic_demod_all).
% 1.89/2.11 dependent: set(dynamic_demod).
% 1.89/2.11 dependent: set(order_eq).
% 1.89/2.11 dependent: set(back_demod).
% 1.89/2.11 dependent: set(lrpo).
% 1.89/2.11 dependent: set(hyper_res).
% 1.89/2.11 dependent: clear(order_hyper).
% 1.89/2.11
% 1.89/2.11 ------------> process usable:
% 1.89/2.11 ** KEPT (pick-wt=5): 2 [copy,1,flip.1] cons(A,B)!=nil.
% 1.89/2.11 ** KEPT (pick-wt=4): 3 [] s(A)!=z.
% 1.89/2.11 ** KEPT (pick-wt=10): 4 [] drop(A,B)!=drop(A,C)|B=C.
% 1.89/2.11
% 1.89/2.11 ------------> process sos:
% 1.89/2.11 ** KEPT (pick-wt=3): 5 [] A=A.
% 1.89/2.11 ** KEPT (pick-wt=6): 6 [] head(cons(A,B))=A.
% 1.89/2.11 ---> New Demodulator: 7 [new_demod,6] head(cons(A,B))=A.
% 1.89/2.11 ** KEPT (pick-wt=6): 8 [] tail(cons(A,B))=B.
% 1.89/2.11 ---> New Demodulator: 9 [new_demod,8] tail(cons(A,B))=B.
% 1.89/2.11 ** KEPT (pick-wt=5): 10 [] proj1S(s(A))=A.
% 1.89/2.11 ---> New Demodulator: 11 [new_demod,10] proj1S(s(A))=A.
% 1.89/2.11 ** KEPT (pick-wt=6): 12 [] drop(s(A),nil)=nil.
% 1.89/2.11 ---> New Demodulator: 13 [new_demod,12] drop(s(A),nil)=nil.
% 1.89/2.11 ** KEPT (pick-wt=10): 14 [] drop(s(A),cons(B,C))=drop(A,C).
% 1.89/2.11 ---> New Demodulator: 15 [new_demod,14] drop(s(A),cons(B,C))=drop(A,C).
% 1.89/2.11 ** KEPT (pick-wt=5): 16 [] drop(z,A)=A.
% 1.89/2.11 ---> New Demodulator: 17 [new_demod,16] drop(z,A)=A.
% 1.89/2.11 Following clause subsumed by 5 during input processing: 0 [copy,5,flip.1] A=A.
% 1.89/2.11 >>>> Starting back demodulation with 7.
% 1.89/2.11 >>>> Starting back demodulation with 9.
% 1.89/2.11 >>>> Starting back demodulation with 11.
% 1.89/2.11 >>>> Starting back demodulation with 13.
% 1.89/2.11 >>>> Starting back demodulation with 15.
% 1.89/2.11 >>>> Starting back demodulation with 17.
% 1.89/2.11
% 1.89/2.11 ======= end of input processing =======
% 1.89/2.11
% 1.89/2.11 =========== start of search ===========
% 1.89/2.11
% 1.89/2.11 -------- PROOF --------
% 1.89/2.11
% 1.89/2.11 -----> EMPTY CLAUSE at 0.02 sec ----> 255 [hyper,254,5,16] $F.
% 1.89/2.11
% 1.89/2.11 Length of proof is 4. Level of proof is 3.
% 1.89/2.11
% 1.89/2.11 ---------------- PROOF ----------------
% 1.89/2.11 % SZS status Theorem
% 1.89/2.11 % SZS output start Refutation
% See solution above
% 1.89/2.11 ------------ end of proof -------------
% 1.89/2.11
% 1.89/2.11
% 1.89/2.11 Search stopped by max_proofs option.
% 1.89/2.11
% 1.89/2.11
% 1.89/2.11 Search stopped by max_proofs option.
% 1.89/2.11
% 1.89/2.11 ============ end of search ============
% 1.89/2.11
% 1.89/2.11 -------------- statistics -------------
% 1.89/2.11 clauses given 23
% 1.89/2.11 clauses generated 376
% 1.89/2.11 clauses kept 247
% 1.89/2.11 clauses forward subsumed 125
% 1.89/2.11 clauses back subsumed 3
% 1.89/2.11 Kbytes malloced 976
% 1.89/2.11
% 1.89/2.11 ----------- times (seconds) -----------
% 1.89/2.11 user CPU time 0.02 (0 hr, 0 min, 0 sec)
% 1.89/2.11 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.89/2.11 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.89/2.11
% 1.89/2.11 That finishes the proof of the theorem.
% 1.89/2.11
% 1.89/2.11 Process 22345 finished Tue May 5 09:37:33 2026
% 1.89/2.11 Otter interrupted
% 1.89/2.11 PROOF FOUND
%------------------------------------------------------------------------------