YES Problem: 2nd(cons(X,n__cons(Y,Z))) -> activate(Y) from(X) -> cons(X,n__from(s(X))) cons(X1,X2) -> n__cons(X1,X2) from(X) -> n__from(X) activate(n__cons(X1,X2)) -> cons(X1,X2) activate(n__from(X)) -> from(X) activate(X) -> X Proof: DP Processor: DPs: 2nd#(cons(X,n__cons(Y,Z))) -> activate#(Y) from#(X) -> cons#(X,n__from(s(X))) activate#(n__cons(X1,X2)) -> cons#(X1,X2) activate#(n__from(X)) -> from#(X) TRS: 2nd(cons(X,n__cons(Y,Z))) -> activate(Y) from(X) -> cons(X,n__from(s(X))) cons(X1,X2) -> n__cons(X1,X2) from(X) -> n__from(X) activate(n__cons(X1,X2)) -> cons(X1,X2) activate(n__from(X)) -> from(X) activate(X) -> X Arctic Interpretation Processor: dimension: 1 interpretation: [cons#](x0, x1) = 2, [from#](x0) = 4, [activate#](x0) = x0 + 5, [2nd#](x0) = x0 + 8, [n__from](x0) = 2x0, [s](x0) = x0 + 2, [from](x0) = 4x0 + 6, [activate](x0) = 3x0 + 8, [2nd](x0) = 1x0 + 13, [cons](x0, x1) = 3x0 + 1x1 + 5, [n__cons](x0, x1) = 2x0 + x1 + 4 orientation: 2nd#(cons(X,n__cons(Y,Z))) = 3X + 3Y + 1Z + 8 >= Y + 5 = activate#(Y) from#(X) = 4 >= 2 = cons#(X,n__from(s(X))) activate#(n__cons(X1,X2)) = 2X1 + X2 + 5 >= 2 = cons#(X1,X2) activate#(n__from(X)) = 2X + 5 >= 4 = from#(X) 2nd(cons(X,n__cons(Y,Z))) = 4X + 4Y + 2Z + 13 >= 3Y + 8 = activate(Y) from(X) = 4X + 6 >= 3X + 5 = cons(X,n__from(s(X))) cons(X1,X2) = 3X1 + 1X2 + 5 >= 2X1 + X2 + 4 = n__cons(X1,X2) from(X) = 4X + 6 >= 2X = n__from(X) activate(n__cons(X1,X2)) = 5X1 + 3X2 + 8 >= 3X1 + 1X2 + 5 = cons(X1,X2) activate(n__from(X)) = 5X + 8 >= 4X + 6 = from(X) activate(X) = 3X + 8 >= X = X problem: DPs: TRS: 2nd(cons(X,n__cons(Y,Z))) -> activate(Y) from(X) -> cons(X,n__from(s(X))) cons(X1,X2) -> n__cons(X1,X2) from(X) -> n__from(X) activate(n__cons(X1,X2)) -> cons(X1,X2) activate(n__from(X)) -> from(X) activate(X) -> X Qed