YES (ignored inputs)COMMENT experiments for [36] submitted by: Takahito Aoto Rewrite Rules: [ +(0,?y) -> ?y, +(s(0),?y) -> s(+(0,?y)), +(?x,?y) -> +(?y,?x), s(s(?x)) -> ?x ] Apply Direct Methods... Inner CPs: [ s(?x) = s(?x) ] Outer CPs: [ ?y = +(?y,0), s(+(0,?y_1)) = +(?y_1,s(0)) ] not Overlay, check Termination... unknown/not Terminating unknown Knuth & Bendix Linear unknown Development Closed unknown Strongly Closed unknown Weakly-Non-Overlapping & Non-Collapsing & Shallow unknown Upside-Parallel-Closed/Outside-Closed (inner) Parallel CPs: (not computed) unknown Toyama (Parallel CPs) Simultaneous CPs: [ +(?y,0) = ?y, +(?y,s(0)) = s(+(0,?y)), ?y = +(?y,0), s(+(0,?y)) = +(?y,s(0)), s(?x_1) = s(?x_1), ?x_1 = s(s(?x_1)) ] unknown Okui (Simultaneous CPs) unknown Strongly Depth-Preserving & Root-E-Closed/Non-E-Overlapping unknown Strongly Weight-Preserving & Root-E-Closed/Non-E-Overlapping check Locally Decreasing Diagrams by Rule Labelling... Critical Pair by Rules <3, 3> preceded by [(s,1)] joinable by a reduction of rules <[], []> Critical Pair <+(?y_2,0), ?y_2> by Rules <2, 0> preceded by [] joinable by a reduction of rules <[([],2),([],0)], []> Critical Pair <+(?y_2,s(0)), s(+(0,?y_2))> by Rules <2, 1> preceded by [] joinable by a reduction of rules <[([],2),([],1)], []> unknown Diagram Decreasing check Non-Confluence... obtain 7 rules by 3 steps unfolding obtain 24 candidates for checking non-joinability check by TCAP-Approximation (failure) check by Ordering(rpo), check by Tree-Automata Approximation (failure) check by Interpretation(mod2) (failure) check by Descendants-Approximation, check by Ordering(poly) (failure) unknown Non-Confluence Check relative termination: [ +(0,?y) -> ?y, +(s(0),?y) -> s(+(0,?y)), s(s(?x)) -> ?x ] [ +(?x,?y) -> +(?y,?x) ] Polynomial Interpretation: +:= (2)*x1+(1)*x1*x1+(2)*x2+(1)*x2*x2 0:= 0 s:= (4)+(1)*x1 retract +(s(0),?y) -> s(+(0,?y)) retract s(s(?x)) -> ?x Polynomial Interpretation: +:= (2)+(1)*x1+(1)*x2 0:= 0 s:= (4)*x1 relatively terminating unknown Huet (modulo AC) check by Reduction-Preserving Completion... STEP: 1 (parallel) S: [ +(0,?y) -> ?y, +(s(0),?y) -> s(+(0,?y)), s(s(?x)) -> ?x ] P: [ +(?x,?y) -> +(?y,?x) ] S: terminating CP(S,S): --> => yes PCP_in(symP,S): CP(S,symP): --> => no --> => no check joinability condition: check modulo reachablity from ?y to +(?y,0): maybe not reachable check modulo reachablity from s(?y) to +(?y,s(0)): maybe not reachable failed failure(Step 1) [ +(?y,0) -> ?y, +(?y,s(0)) -> s(?y) ] Added S-Rules: [ +(?y,0) -> ?y, +(?y,s(0)) -> s(?y) ] Added P-Rules: [ ] replace: +(s(0),?y) -> s(+(0,?y)) => +(s(0),?y) -> s(+(?y,0)) STEP: 2 (linear) S: [ +(0,?y) -> ?y, +(s(0),?y) -> s(+(0,?y)), s(s(?x)) -> ?x ] P: [ +(?x,?y) -> +(?y,?x) ] S: terminating CP(S,S): --> => yes CP_in(symP,S): CP(S,symP): --> => no --> => no check joinability condition: check modulo reachablity from ?y to +(?y,0): maybe not reachable check modulo reachablity from s(?y) to +(?y,s(0)): maybe not reachable failed failure(Step 2) [ +(?y,0) -> ?y, +(?y,s(0)) -> s(?y) ] Added S-Rules: [ +(?y,0) -> ?y, +(?y,s(0)) -> s(?y) ] Added P-Rules: [ ] replace: +(s(0),?y) -> s(+(0,?y)) => +(s(0),?y) -> s(+(?y,0)) STEP: 3 (relative) S: [ +(0,?y) -> ?y, +(s(0),?y) -> s(+(0,?y)), s(s(?x)) -> ?x ] P: [ +(?x,?y) -> +(?y,?x) ] Check relative termination: [ +(0,?y) -> ?y, +(s(0),?y) -> s(+(0,?y)), s(s(?x)) -> ?x ] [ +(?x,?y) -> +(?y,?x) ] Polynomial Interpretation: +:= (2)*x1+(1)*x1*x1+(2)*x2+(1)*x2*x2 0:= 0 s:= (4)+(1)*x1 retract +(s(0),?y) -> s(+(0,?y)) retract s(s(?x)) -> ?x Polynomial Interpretation: +:= (2)+(1)*x1+(1)*x2 0:= 0 s:= (4)*x1 relatively terminating S/P: relatively terminating check CP condition: failed failure(Step 3) STEP: 4 (parallel) S: [ +(0,?y) -> ?y, +(s(0),?y) -> s(+(0,?y)), s(s(?x)) -> ?x, +(?y,0) -> ?y, +(?y,s(0)) -> s(?y) ] P: [ +(?x,?y) -> +(?y,?x) ] S: terminating CP(S,S): <0, 0> --> <0, 0> => yes --> => yes --> => yes --> <0, 0> => yes --> => yes <0, 0> --> <0, 0> => yes --> => yes --> => yes --> <0, 0> => yes PCP_in(symP,S): CP(S,symP): --> => yes --> => yes --> => yes --> => yes S: [ +(0,?y) -> ?y, +(s(0),?y) -> s(+(0,?y)), s(s(?x)) -> ?x, +(?y,0) -> ?y, +(?y,s(0)) -> s(?y) ] P: [ +(?x,?y) -> +(?y,?x) ] Success Reduction-Preserving Completion Direct Methods: CR Combined result: CR 581.trs: Success(CR) (310 msec.)