YES Problem 1: (VAR v_NonEmpty:S x:S y:S) (RULES if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: IF_MINUS(ffalse,s(x:S),y:S) -> MINUS(x:S,y:S) LE(s(x:S),s(y:S)) -> LE(x:S,y:S) LOG(s(s(x:S))) -> LOG(s(quot(x:S,s(s(0))))) LOG(s(s(x:S))) -> QUOT(x:S,s(s(0))) MINUS(s(x:S),y:S) -> IF_MINUS(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: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: IF_MINUS(ffalse,s(x:S),y:S) -> MINUS(x:S,y:S) LE(s(x:S),s(y:S)) -> LE(x:S,y:S) LOG(s(s(x:S))) -> LOG(s(quot(x:S,s(s(0))))) LOG(s(s(x:S))) -> QUOT(x:S,s(s(0))) MINUS(s(x:S),y:S) -> IF_MINUS(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: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: IF_MINUS(ffalse,s(x:S),y:S) -> MINUS(x:S,y:S) MINUS(s(x:S),y:S) -> IF_MINUS(le(s(x:S),y:S),s(x:S),y:S) ->->-> Rules: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: LOG(s(s(x:S))) -> LOG(s(quot(x:S,s(s(0))))) ->->-> Rules: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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 4 subproblems. Problem 1.1: Subterm Processor: -> Pairs: LE(s(x:S),s(y:S)) -> LE(x:S,y:S) -> Rules: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: IF_MINUS(ffalse,s(x:S),y:S) -> MINUS(x:S,y:S) MINUS(s(x:S),y:S) -> IF_MINUS(le(s(x:S),y:S),s(x:S),y:S) -> Rules: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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(IF_MINUS) = 2 pi(MINUS) = 1 Problem 1.2: SCC Processor: -> Pairs: MINUS(s(x:S),y:S) -> IF_MINUS(le(s(x:S),y:S),s(x:S),y:S) -> Rules: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) -> if_minus(le(s(x:S),y:S),s(x:S),y:S) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [if_minus](X1,X2,X3) = 2.X2 + 1 [le](X1,X2) = 2.X2 + 2 [log](X) = 0 [minus](X1,X2) = 2.X1 + 1 [quot](X1,X2) = 0 [0] = 0 [fSNonEmpty] = 0 [false] = 2 [s](X) = 2.X + 2 [true] = 1 [IF_MINUS](X1,X2,X3) = 0 [LE](X1,X2) = 0 [LOG](X) = 0 [MINUS](X1,X2) = 0 [QUOT](X1,X2) = X1 Problem 1.3: SCC Processor: -> Pairs: Empty -> Rules: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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 1.4: Reduction Pairs Processor: -> Pairs: LOG(s(s(x:S))) -> LOG(s(quot(x:S,s(s(0))))) -> Rules: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) -> if_minus(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))) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [if_minus](X1,X2,X3) = X2 [le](X1,X2) = 2.X1 + 2 [log](X) = 0 [minus](X1,X2) = X1 [quot](X1,X2) = 2.X1 [0] = 0 [fSNonEmpty] = 0 [false] = 2 [s](X) = 2.X + 2 [true] = 2 [IF_MINUS](X1,X2,X3) = 0 [LE](X1,X2) = 0 [LOG](X) = 2.X [MINUS](X1,X2) = 0 [QUOT](X1,X2) = 0 Problem 1.4: SCC Processor: -> Pairs: Empty -> Rules: if_minus(ffalse,s(x:S),y:S) -> s(minus(x:S,y:S)) if_minus(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) log(s(0)) -> 0 log(s(s(x:S))) -> s(log(s(quot(x:S,s(s(0)))))) minus(0,y:S) -> 0 minus(s(x:S),y:S) -> if_minus(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.