YES Problem: +(0(),y) -> y +(x,0()) -> x +(s(x),y) -> s(+(x,y)) +(x,s(y)) -> s(+(x,y)) +(x,+(y,z)) -> +(+(x,y),z) +(x,y) -> +(y,x) Proof: AT confluence processor Complete TRS T' of input TRS: +(0(),y) -> y +(x,0()) -> x +(s(x),y) -> s(+(x,y)) +(x,s(y)) -> s(+(x,y)) +(x,+(y,z)) -> +(+(x,y),z) +(x,y) -> +(y,x) T' = (P union S) with TRS P:+(x,+(y,z)) -> +(+(x,y),z) +(x,y) -> +(y,x) TRS S:+(0(),y) -> y +(x,0()) -> x +(s(x),y) -> s(+(x,y)) +(x,s(y)) -> s(+(x,y)) S is left-linear and P is reversible. CP(S,S) = 0() = 0(), s(y) = s(+(0(),y)), s(x) = s(+(x,0())), s(+(x221,0())) = s(x221), s(+(x223,s(y))) = s(+(s(x223),y)), s(+(0(),x226)) = s(x226), s(+(s(x),x228)) = s(+(x,s(x228))) CP(S,P union P^-1) = +(y,z) = +(+(0(),y),z), +(x,z) = +(+(x,0()),z), y = +(y,0()), +(y,z) = +(0(),+(y,z)), x = +(x,0()), +(x,y) = +(+(x,y),0()), x = +(0(),x), +(x,y) = +(x,+(y,0())), +(x,z) = +(x,+(0(),z)), y = +(0(),y), s(+(x275,+(y,z))) = +(+(s(x275),y),z), +(x,s(+(x277,z))) = +(+(x,s(x277)),z), s(+(x279,y)) = +(y,s(x279)), +(s(+(x281,y)),z) = +(s(x281),+(y,z)), s(+(x283,x)) = +(x,s(x283)), +(x,s(+(y,x286))) = +(+(x,y),s(x286)), s(+(x,x288)) = +(s(x288),x), s(+(+(x,y),x290)) = +(x,+(y,s(x290))), +(s(+(x,x292)),z) = +(x,+(s(x292),z)), s(+(y,x294)) = +(s(x294),y) PCP_in(P union P^-1,S) = We have to check termination of S: Matrix Interpretation Processor: dim=1 interpretation: [+](x0, x1) = x0 + x1 + 6, [0] = 1, [s](x0) = x0 + 1 orientation: +(0(),y) = y + 7 >= y = y +(x,0()) = x + 7 >= x = x +(s(x),y) = x + y + 7 >= x + y + 7 = s(+(x,y)) +(x,s(y)) = x + y + 7 >= x + y + 7 = s(+(x,y)) problem: +(s(x),y) -> s(+(x,y)) +(x,s(y)) -> s(+(x,y)) Matrix Interpretation Processor: dim=1 interpretation: [+](x0, x1) = x0 + 2x1 + 7, [s](x0) = x0 + 6 orientation: +(s(x),y) = x + 2y + 13 >= x + 2y + 13 = s(+(x,y)) +(x,s(y)) = x + 2y + 19 >= x + 2y + 13 = s(+(x,y)) problem: +(s(x),y) -> s(+(x,y)) Matrix Interpretation Processor: dim=1 interpretation: [+](x0, x1) = 6x0 + x1 + 2, [s](x0) = x0 + 3 orientation: +(s(x),y) = 6x + y + 20 >= 6x + y + 5 = s(+(x,y)) problem: Qed