/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 minus(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(minus(x,y),z) -> MINUS(x,plus(y,z)) MINUS(minus(x,y),z) -> PLUS(y,z) MINUS(s(x),s(y)) -> MINUS(x,y) 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(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(minus(x,y),z) -> MINUS(x,plus(y,z)) MINUS(minus(x,y),z) -> PLUS(y,z) MINUS(s(x),s(y)) -> MINUS(x,y) 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(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(s(x),y) -> PLUS(x,y) ->->-> Rules: minus(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(minus(x,y),z) -> MINUS(x,plus(y,z)) MINUS(s(x),s(y)) -> MINUS(x,y) ->->-> Rules: minus(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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: Subterm Processor: -> Pairs: PLUS(s(x),y) -> PLUS(x,y) -> Rules: minus(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(PLUS) = 1 Problem 1.1: SCC Processor: -> Pairs: Empty -> Rules: minus(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(minus(x,y),z) -> MINUS(x,plus(y,z)) MINUS(s(x),s(y)) -> MINUS(x,y) -> Rules: minus(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x plus(0,y) -> y plus(s(x),y) -> s(plus(x,y)) ->Interpretation type: Linear ->Coefficients: Natural Numbers ->Dimension: 1 ->Bound: 2 ->Interpretation: [minus](X1,X2) = X1 + 1 [plus](X1,X2) = 2.X1 + 2.X2 + 2 [0] = 0 [s](X) = X + 2 [QUOT](X1,X2) = 2.X1 Problem 1.3: SCC Processor: -> Pairs: Empty -> Rules: minus(minus(x,y),z) -> minus(x,plus(y,z)) minus(s(x),s(y)) -> minus(x,y) minus(x,0) -> x 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.