----- Otter 3.2, August 2001 ----- The process was started by ??? on ???, Sat Nov 22 00:15:16 2003 The command was "otter". set(auto). dependent: set(auto1). dependent: set(process_input). dependent: clear(print_kept). dependent: clear(print_new_demod). dependent: clear(print_back_demod). dependent: clear(print_back_sub). dependent: set(control_memory). dependent: assign(max_mem, 12000). dependent: assign(pick_given_ratio, 4). dependent: assign(stats_level, 1). dependent: assign(max_seconds, 10800). clear(print_given). assign(max_literals,1). include("booleanAx.txt"). ------- start included file booleanAx.txt------- include("1_2booleanAx.txt"). ------- start included file 1_2booleanAx.txt------- formula_list(usable). all x y (u(x,y)=u(y,x)). all x y z (u(x,u(y,z))=u(u(x,y),z)). end_of_list. -------> usable clausifies to: list(usable). 0 [] u(x,y)=u(y,x). 0 [] u(x,u(y,z))=u(u(x,y),z). end_of_list. ------- end included file 1_2booleanAx.txt------- include("3booleanAx.txt"). ------- start included file 3booleanAx.txt------- formula_list(usable). all x y (u(c(u(c(x),y)),c(u(c(x),c(y))))=x). end_of_list. -------> usable clausifies to: list(usable). 0 [] u(c(u(c(x),y)),c(u(c(x),c(y))))=x. end_of_list. ------- end included file 3booleanAx.txt------- ------- end included file booleanAx.txt------- include("peirceanAx0.txt"). ------- start included file peirceanAx0.txt------- formula_list(usable). all x y z (k(x,k(y,z))=k(k(x,y),z)). all x (r(r(x))=x). all x y (r(u(x,y))=u(r(x),r(y))). all x y (r(k(x,y))=k(r(y),r(x))). end_of_list. -------> usable clausifies to: list(usable). 0 [] k(x,k(y,z))=k(k(x,y),z). 0 [] r(r(x))=x. 0 [] r(u(x,y))=u(r(x),r(y)). 0 [] r(k(x,y))=k(r(y),r(x)). end_of_list. ------- end included file peirceanAx0.txt------- include("peirceanAx2.txt"). ------- start included file peirceanAx2.txt------- formula_list(usable). all x (k(x,I)=x). all x y (u(k(r(x),c(k(x,y))),c(y))=c(y)). end_of_list. -------> usable clausifies to: list(usable). 0 [] k(x,I)=x. 0 [] u(k(r(x),c(k(x,y))),c(y))=c(y). end_of_list. ------- end included file peirceanAx2.txt------- include("iDef.txt"). ------- start included file iDef.txt------- formula_list(usable). all x y (i(x,y)<->u(x,y)=y). end_of_list. -------> usable clausifies to: list(usable). 0 [] -i(x,y)|u(x,y)=y. 0 [] i(x,y)|u(x,y)!=y. end_of_list. ------- end included file iDef.txt------- include("uzDef.txt"). ------- start included file uzDef.txt------- formula_list(usable). D=c(I). U=u(I,D). Z=c(U). end_of_list. -------> usable clausifies to: list(usable). 0 [] D=c(I). 0 [] U=u(I,D). 0 [] Z=c(U). end_of_list. ------- end included file uzDef.txt------- include("dDef.txt"). ------- start included file dDef.txt------- formula_list(usable). all x y (d(x,y)=c(k(c(x),c(y)))). end_of_list. -------> usable clausifies to: list(usable). 0 [] d(x,y)=c(k(c(x),c(y))). end_of_list. ------- end included file dDef.txt------- include("xi_a.txt"). ------- start included file xi_a.txt------- formula_list(usable). -(all x i(k(r(x),c(x)),D)). end_of_list. -------> usable clausifies to: list(usable). 0 [] -i(k(r($c1),c($c1)),D). end_of_list. ------- end included file xi_a.txt------- SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=2. This is a Horn set with equality. The strategy will be Knuth-Bendix and hyper_res, with positive clauses in sos and nonpositive clauses in usable. dependent: set(knuth_bendix). dependent: set(para_from). dependent: set(para_into). dependent: clear(para_from_right). dependent: clear(para_into_right). dependent: set(para_from_vars). dependent: set(eq_units_both_ways). dependent: set(dynamic_demod_all). dependent: set(dynamic_demod). dependent: set(order_eq). dependent: set(back_demod). dependent: set(lrpo). dependent: set(hyper_res). dependent: clear(order_hyper). ------------> process usable: ** KEPT (pick-wt=8): 1 [] -i(x,y)|u(x,y)=y. ** KEPT (pick-wt=8): 2 [] i(x,y)|u(x,y)!=y. ** KEPT (pick-wt=7): 3 [] -i(k(r($c1),c($c1)),D). ------------> process sos: ** KEPT (pick-wt=7): 4 [] u(x,y)=u(y,x). ** KEPT (pick-wt=11): 6 [copy,5,flip.1] u(u(x,y),z)=u(x,u(y,z)). ---> New Demodulator: 7 [new_demod,6] u(u(x,y),z)=u(x,u(y,z)). ** KEPT (pick-wt=14): 8 [] u(c(u(c(x),y)),c(u(c(x),c(y))))=x. ---> New Demodulator: 9 [new_demod,8] u(c(u(c(x),y)),c(u(c(x),c(y))))=x. ** KEPT (pick-wt=11): 11 [copy,10,flip.1] k(k(x,y),z)=k(x,k(y,z)). ---> New Demodulator: 12 [new_demod,11] k(k(x,y),z)=k(x,k(y,z)). ** KEPT (pick-wt=5): 13 [] r(r(x))=x. ---> New Demodulator: 14 [new_demod,13] r(r(x))=x. ** KEPT (pick-wt=10): 15 [] r(u(x,y))=u(r(x),r(y)). ---> New Demodulator: 16 [new_demod,15] r(u(x,y))=u(r(x),r(y)). ** KEPT (pick-wt=10): 17 [] r(k(x,y))=k(r(y),r(x)). ---> New Demodulator: 18 [new_demod,17] r(k(x,y))=k(r(y),r(x)). ** KEPT (pick-wt=5): 19 [] k(x,I)=x. ---> New Demodulator: 20 [new_demod,19] k(x,I)=x. ** KEPT (pick-wt=13): 21 [] u(k(r(x),c(k(x,y))),c(y))=c(y). ---> New Demodulator: 22 [new_demod,21] u(k(r(x),c(k(x,y))),c(y))=c(y). ** KEPT (pick-wt=4): 24 [copy,23,flip.1] c(I)=D. ---> New Demodulator: 25 [new_demod,24] c(I)=D. ** KEPT (pick-wt=5): 27 [copy,26,flip.1] u(I,D)=U. ---> New Demodulator: 28 [new_demod,27] u(I,D)=U. ** KEPT (pick-wt=4): 30 [copy,29,flip.1] c(U)=Z. ---> New Demodulator: 31 [new_demod,30] c(U)=Z. ** KEPT (pick-wt=10): 33 [copy,32,flip.1] c(k(c(x),c(y)))=d(x,y). ---> New Demodulator: 34 [new_demod,33] c(k(c(x),c(y)))=d(x,y). Following clause subsumed by 4 during input processing: 0 [copy,4,flip.1] u(x,y)=u(y,x). >>>> Starting back demodulation with 7. >>>> Starting back demodulation with 9. >>>> Starting back demodulation with 12. >>>> Starting back demodulation with 14. >>>> Starting back demodulation with 16. >>>> Starting back demodulation with 18. >>>> Starting back demodulation with 20. >>>> Starting back demodulation with 22. >>>> Starting back demodulation with 25. >>>> Starting back demodulation with 28. >>>> Starting back demodulation with 31. >>>> Starting back demodulation with 34. ======= end of input processing ======= =========== start of search =========== ----> UNIT CONFLICT at 0.04 sec ----> 284 [binary,283.1,3.1] $F. Length of proof is 3. Level of proof is 3. ---------------- PROOF ---------------- 2 [] i(x,y)|u(x,y)!=y. 3 [] -i(k(r($c1),c($c1)),D). 19 [] k(x,I)=x. 21 [] u(k(r(x),c(k(x,y))),c(y))=c(y). 23 [] D=c(I). 25,24 [copy,23,flip.1] c(I)=D. 152 [para_into,21.1.1.1.2.1,19.1.1,demod,25,25] u(k(r(x),c(x)),D)=D. 283 [hyper,152,2] i(k(r(x),c(x)),D). 284 [binary,283.1,3.1] $F. ------------ end of proof ------------- Search stopped by max_proofs option. ============ end of search ============ -------------- statistics ------------- clauses given 49 clauses generated 490 clauses kept 188 clauses forward subsumed 346 clauses back subsumed 22 Kbytes malloced 287 ----------- times (seconds) ----------- user CPU time 0.24 (0 hr, 0 min, 0 sec) system CPU time 0.00 (0 hr, 0 min, 0 sec) wall-clock time 0 (0 hr, 0 min, 0 sec) hyper_res time 0.00 para_into time 0.00 para_from time 0.00 for_sub time 0.00 back_sub time 0.00 conflict time 0.00 demod time 0.00 That finishes the proof of the theorem. Process 0 finished Sat Nov 22 00:15:16 2003