YES Problem: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) minus(minus(x,y),z) -> minus(x,plus(y,z)) Proof: DP Processor: DPs: minus#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) plus#(s(x),y) -> plus#(x,y) minus#(minus(x,y),z) -> plus#(y,z) minus#(minus(x,y),z) -> minus#(x,plus(y,z)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) minus(minus(x,y),z) -> minus(x,plus(y,z)) TDG Processor: DPs: minus#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) plus#(s(x),y) -> plus#(x,y) minus#(minus(x,y),z) -> plus#(y,z) minus#(minus(x,y),z) -> minus#(x,plus(y,z)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) minus(minus(x,y),z) -> minus(x,plus(y,z)) graph: plus#(s(x),y) -> plus#(x,y) -> plus#(s(x),y) -> plus#(x,y) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) -> quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) -> quot#(s(x),s(y)) -> minus#(x,y) quot#(s(x),s(y)) -> minus#(x,y) -> minus#(minus(x,y),z) -> minus#(x,plus(y,z)) quot#(s(x),s(y)) -> minus#(x,y) -> minus#(minus(x,y),z) -> plus#(y,z) quot#(s(x),s(y)) -> minus#(x,y) -> minus#(s(x),s(y)) -> minus#(x,y) minus#(s(x),s(y)) -> minus#(x,y) -> minus#(minus(x,y),z) -> minus#(x,plus(y,z)) minus#(s(x),s(y)) -> minus#(x,y) -> minus#(minus(x,y),z) -> plus#(y,z) minus#(s(x),s(y)) -> minus#(x,y) -> minus#(s(x),s(y)) -> minus#(x,y) minus#(minus(x,y),z) -> plus#(y,z) -> plus#(s(x),y) -> plus#(x,y) minus#(minus(x,y),z) -> minus#(x,plus(y,z)) -> minus#(minus(x,y),z) -> minus#(x,plus(y,z)) minus#(minus(x,y),z) -> minus#(x,plus(y,z)) -> minus#(minus(x,y),z) -> plus#(y,z) minus#(minus(x,y),z) -> minus#(x,plus(y,z)) -> minus#(s(x),s(y)) -> minus#(x,y) SCC Processor: #sccs: 3 #rules: 4 #arcs: 13/36 DPs: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) minus(minus(x,y),z) -> minus(x,plus(y,z)) Usable Rule Processor: DPs: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) minus(minus(x,y),z) -> minus(x,plus(y,z)) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) Arctic Interpretation Processor: dimension: 1 usable rules: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) minus(minus(x,y),z) -> minus(x,plus(y,z)) interpretation: [quot#](x0, x1) = x0, [plus](x0, x1) = -4x0 + 1x1 + -2, [s](x0) = 1x0 + 0, [minus](x0, x1) = x0 + -4, [0] = 4 orientation: quot#(s(x),s(y)) = 1x + 0 >= x + -4 = quot#(minus(x,y),s(y)) minus(x,0()) = x + -4 >= x = x minus(s(x),s(y)) = 1x + 0 >= x + -4 = minus(x,y) minus(minus(x,y),z) = x + -4 >= x + -4 = minus(x,plus(y,z)) plus(0(),y) = 1y + 0 >= y = y plus(s(x),y) = -3x + 1y + -2 >= -3x + 2y + 0 = s(plus(x,y)) problem: DPs: TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) minus(minus(x,y),z) -> minus(x,plus(y,z)) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) Qed DPs: minus#(s(x),s(y)) -> minus#(x,y) minus#(minus(x,y),z) -> minus#(x,plus(y,z)) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) minus(minus(x,y),z) -> minus(x,plus(y,z)) Subterm Criterion Processor: simple projection: pi(minus#) = 0 problem: DPs: TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) minus(minus(x,y),z) -> minus(x,plus(y,z)) Qed DPs: plus#(s(x),y) -> plus#(x,y) TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) minus(minus(x,y),z) -> minus(x,plus(y,z)) Subterm Criterion Processor: simple projection: pi(plus#) = 0 problem: DPs: TRS: minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) quot(0(),s(y)) -> 0() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) plus(0(),y) -> y plus(s(x),y) -> s(plus(x,y)) minus(minus(x,y),z) -> minus(x,plus(y,z)) Qed