YES Problem: le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) mod(0(),y) -> 0() mod(s(x),0()) -> 0() mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) if_mod(false(),s(x),s(y)) -> s(x) Proof: DP Processor: DPs: le#(s(x),s(y)) -> le#(x,y) minus#(s(x),s(y)) -> minus#(x,y) mod#(s(x),s(y)) -> le#(y,x) mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y)) if_mod#(true(),s(x),s(y)) -> minus#(x,y) if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y)) TRS: le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) mod(0(),y) -> 0() mod(s(x),0()) -> 0() mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y)) if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y)) if_mod(false(),s(x),s(y)) -> s(x) Usable Rule Processor: DPs: le#(s(x),s(y)) -> le#(x,y) minus#(s(x),s(y)) -> minus#(x,y) mod#(s(x),s(y)) -> le#(y,x) mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y)) if_mod#(true(),s(x),s(y)) -> minus#(x,y) if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y)) TRS: le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) Arctic Interpretation Processor: dimension: 1 usable rules: le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) interpretation: [if_mod#](x0, x1, x2) = 2x0 + x1 + 0, [mod#](x0, x1) = 2x0 + 0, [minus#](x0, x1) = 1x0, [le#](x0, x1) = x1, [minus](x0, x1) = 1x0, [false] = 4, [s](x0) = 4x0 + 3, [true] = 0, [le](x0, x1) = 3x1 + 1, [0] = 2 orientation: le#(s(x),s(y)) = 4y + 3 >= y = le#(x,y) minus#(s(x),s(y)) = 5x + 4 >= 1x = minus#(x,y) mod#(s(x),s(y)) = 6x + 5 >= x = le#(y,x) mod#(s(x),s(y)) = 6x + 5 >= 5x + 3 = if_mod#(le(y,x),s(x),s(y)) if_mod#(true(),s(x),s(y)) = 4x + 3 >= 1x = minus#(x,y) if_mod#(true(),s(x),s(y)) = 4x + 3 >= 3x + 0 = mod#(minus(x,y),s(y)) le(0(),y) = 3y + 1 >= 0 = true() le(s(x),0()) = 5 >= 4 = false() le(s(x),s(y)) = 7y + 6 >= 3y + 1 = le(x,y) minus(x,0()) = 1x >= x = x minus(s(x),s(y)) = 5x + 4 >= 1x = minus(x,y) problem: DPs: TRS: le(0(),y) -> true() le(s(x),0()) -> false() le(s(x),s(y)) -> le(x,y) minus(x,0()) -> x minus(s(x),s(y)) -> minus(x,y) Qed