/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR x y z) (RULES plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) ) Problem 1: Innermost Equivalent Processor: -> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: PLUS(s(x),y) -> PLUS(x,y) QUOT(s(x),s(y),z) -> QUOT(x,y,z) QUOT(x,0,s(z)) -> PLUS(z,s(0)) QUOT(x,0,s(z)) -> QUOT(x,plus(z,s(0)),s(z)) -> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) Problem 1: SCC Processor: -> Pairs: PLUS(s(x),y) -> PLUS(x,y) QUOT(s(x),s(y),z) -> QUOT(x,y,z) QUOT(x,0,s(z)) -> PLUS(z,s(0)) QUOT(x,0,s(z)) -> QUOT(x,plus(z,s(0)),s(z)) -> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: PLUS(s(x),y) -> PLUS(x,y) ->->-> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) ->->Cycle: ->->-> Pairs: QUOT(s(x),s(y),z) -> QUOT(x,y,z) QUOT(x,0,s(z)) -> QUOT(x,plus(z,s(0)),s(z)) ->->-> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) The problem is decomposed in 2 subproblems. Problem 1.1: Subterm Processor: -> Pairs: PLUS(s(x),y) -> PLUS(x,y) -> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) ->Projection: pi(PLUS) = 1 Problem 1.1: SCC Processor: -> Pairs: Empty -> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) ->Strongly Connected Components: There is no strongly connected component The problem is finite. Problem 1.2: Subterm Processor: -> Pairs: QUOT(s(x),s(y),z) -> QUOT(x,y,z) QUOT(x,0,s(z)) -> QUOT(x,plus(z,s(0)),s(z)) -> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) ->Projection: pi(QUOT) = 1 Problem 1.2: SCC Processor: -> Pairs: QUOT(x,0,s(z)) -> QUOT(x,plus(z,s(0)),s(z)) -> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: QUOT(x,0,s(z)) -> QUOT(x,plus(z,s(0)),s(z)) ->->-> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) Problem 1.2: Reduction Pairs Processor: -> Pairs: QUOT(x,0,s(z)) -> QUOT(x,plus(z,s(0)),s(z)) -> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) -> Usable rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [plus](X1,X2) = X2 + 1 [0] = 2 [s](X) = 0 [QUOT](X1,X2,X3) = 2.X2 Problem 1.2: SCC Processor: -> Pairs: Empty -> Rules: plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y),s(z)) -> 0 quot(s(x),s(y),z) -> quot(x,y,z) quot(x,0,s(z)) -> s(quot(x,plus(z,s(0)),s(z))) ->Strongly Connected Components: There is no strongly connected component The problem is finite.