/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR v_NonEmpty:S X:S Y:S) (RULES ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ) Problem 1: Innermost Equivalent Processor: -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: IFMINUS(ffalse,s(X:S),Y:S) -> MINUS(X:S,Y:S) LE(s(X:S),s(Y:S)) -> LE(X:S,Y:S) MINUS(s(X:S),Y:S) -> IFMINUS(le(s(X:S),Y:S),s(X:S),Y:S) MINUS(s(X:S),Y:S) -> LE(s(X:S),Y:S) QUOT(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S) QUOT(s(X:S),s(Y:S)) -> QUOT(minus(X:S,Y:S),s(Y:S)) -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) Problem 1: SCC Processor: -> Pairs: IFMINUS(ffalse,s(X:S),Y:S) -> MINUS(X:S,Y:S) LE(s(X:S),s(Y:S)) -> LE(X:S,Y:S) MINUS(s(X:S),Y:S) -> IFMINUS(le(s(X:S),Y:S),s(X:S),Y:S) MINUS(s(X:S),Y:S) -> LE(s(X:S),Y:S) QUOT(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S) QUOT(s(X:S),s(Y:S)) -> QUOT(minus(X:S,Y:S),s(Y:S)) -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: LE(s(X:S),s(Y:S)) -> LE(X:S,Y:S) ->->-> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->->Cycle: ->->-> Pairs: IFMINUS(ffalse,s(X:S),Y:S) -> MINUS(X:S,Y:S) MINUS(s(X:S),Y:S) -> IFMINUS(le(s(X:S),Y:S),s(X:S),Y:S) ->->-> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->->Cycle: ->->-> Pairs: QUOT(s(X:S),s(Y:S)) -> QUOT(minus(X:S,Y:S),s(Y:S)) ->->-> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) The problem is decomposed in 3 subproblems. Problem 1.1: Subterm Processor: -> Pairs: LE(s(X:S),s(Y:S)) -> LE(X:S,Y:S) -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->Projection: pi(LE) = 1 Problem 1.1: SCC Processor: -> Pairs: Empty -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->Strongly Connected Components: There is no strongly connected component The problem is finite. Problem 1.2: Subterm Processor: -> Pairs: IFMINUS(ffalse,s(X:S),Y:S) -> MINUS(X:S,Y:S) MINUS(s(X:S),Y:S) -> IFMINUS(le(s(X:S),Y:S),s(X:S),Y:S) -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->Projection: pi(IFMINUS) = 2 pi(MINUS) = 1 Problem 1.2: SCC Processor: -> Pairs: MINUS(s(X:S),Y:S) -> IFMINUS(le(s(X:S),Y:S),s(X:S),Y:S) -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->Strongly Connected Components: There is no strongly connected component The problem is finite. Problem 1.3: Reduction Pairs Processor: -> Pairs: QUOT(s(X:S),s(Y:S)) -> QUOT(minus(X:S,Y:S),s(Y:S)) -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) -> Usable rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [ifMinus](X1,X2,X3) = 2.X2 [le](X1,X2) = 2.X1 + X2 + 2 [minus](X1,X2) = 2.X1 + 1 [quot](X1,X2) = 0 [0] = 2 [fSNonEmpty] = 0 [false] = 2 [s](X) = 2.X + 2 [true] = 1 [IFMINUS](X1,X2,X3) = 0 [LE](X1,X2) = 0 [MINUS](X1,X2) = 0 [QUOT](X1,X2) = 2.X1 Problem 1.3: SCC Processor: -> Pairs: Empty -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->Strongly Connected Components: There is no strongly connected component The problem is finite.