/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR x y z) (RULES minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) ) Problem 1: Dependency Pairs Processor: -> Pairs: MINUS(s(x),s(y)) -> MINUS(x,y) PLUS(minus(x,s(0)),minus(y,s(s(z)))) -> PLUS(minus(y,s(s(z))),minus(x,s(0))) PLUS(s(x),y) -> PLUS(x,y) QUOT(s(x),s(y)) -> MINUS(x,y) QUOT(s(x),s(y)) -> QUOT(minus(x,y),s(y)) -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) Problem 1: SCC Processor: -> Pairs: MINUS(s(x),s(y)) -> MINUS(x,y) PLUS(minus(x,s(0)),minus(y,s(s(z)))) -> PLUS(minus(y,s(s(z))),minus(x,s(0))) PLUS(s(x),y) -> PLUS(x,y) QUOT(s(x),s(y)) -> MINUS(x,y) QUOT(s(x),s(y)) -> QUOT(minus(x,y),s(y)) -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: PLUS(minus(x,s(0)),minus(y,s(s(z)))) -> PLUS(minus(y,s(s(z))),minus(x,s(0))) PLUS(s(x),y) -> PLUS(x,y) ->->-> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) ->->Cycle: ->->-> Pairs: MINUS(s(x),s(y)) -> MINUS(x,y) ->->-> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) ->->Cycle: ->->-> Pairs: QUOT(s(x),s(y)) -> QUOT(minus(x,y),s(y)) ->->-> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) The problem is decomposed in 3 subproblems. Problem 1.1: Reduction Pair Processor: -> Pairs: PLUS(minus(x,s(0)),minus(y,s(s(z)))) -> PLUS(minus(y,s(s(z))),minus(x,s(0))) PLUS(s(x),y) -> PLUS(x,y) -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) -> Usable rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [minus](X1,X2) = X1 + 2.X2 [0] = 2 [s](X) = 2.X + 2 [PLUS](X1,X2) = X1 + X2 Problem 1.1: SCC Processor: -> Pairs: PLUS(minus(x,s(0)),minus(y,s(s(z)))) -> PLUS(minus(y,s(s(z))),minus(x,s(0))) -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: PLUS(minus(x,s(0)),minus(y,s(s(z)))) -> PLUS(minus(y,s(s(z))),minus(x,s(0))) ->->-> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) Problem 1.1: Reduction Pair Processor: -> Pairs: PLUS(minus(x,s(0)),minus(y,s(s(z)))) -> PLUS(minus(y,s(s(z))),minus(x,s(0))) -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) -> Usable rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 2 ->Bound: 1 ->Interpretation: [minus](X1,X2) = [1 1;1 1].X1 + [0 0;1 0].X2 [0] = [1;0] [s](X) = [1 1;0 0].X + [1;1] [PLUS](X1,X2) = [1 0;1 0].X1 + [0 1;1 0].X2 Problem 1.1: SCC Processor: -> Pairs: Empty -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) ->Strongly Connected Components: There is no strongly connected component The problem is finite. Problem 1.2: Subterm Processor: -> Pairs: MINUS(s(x),s(y)) -> MINUS(x,y) -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) ->Projection: pi(MINUS) = 1 Problem 1.2: SCC Processor: -> Pairs: Empty -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) ->Strongly Connected Components: There is no strongly connected component The problem is finite. Problem 1.3: Reduction Pair Processor: -> Pairs: QUOT(s(x),s(y)) -> QUOT(minus(x,y),s(y)) -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) -> Usable rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [minus](X1,X2) = 2.X1 + 1 [0] = 0 [s](X) = 2.X + 2 [QUOT](X1,X2) = 2.X1 Problem 1.3: SCC Processor: -> Pairs: Empty -> Rules: minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(minus(x,s(0)),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0))) plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) quot(0,s(y)) -> 0 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) ->Strongly Connected Components: There is no strongly connected component The problem is finite.