YES * Step 1: FromIts YES + Considered Problem: Rules: 0. sqrt(A,B,C,D) -> f(0,1,1,D) True (1,1) 1. f(A,B,C,D) -> f(1 + A,2 + B,2 + B + C,D) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -1*B + C >= 0 && -1 + A + C >= 0 && -1 + -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= C && B >= 0] 2. f(A,B,C,D) -> end(A,B,C,D) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -1*B + C >= 0 && -1 + A + C >= 0 && -1 + -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1 + D] Signature: {(end,4);(f,4);(sqrt,4)} Flow Graph: [0->{1,2},1->{1,2},2->{}] + Applied Processor: FromIts + Details: () * Step 2: Decompose YES + Considered Problem: Rules: sqrt(A,B,C,D) -> f(0,1,1,D) True f(A,B,C,D) -> f(1 + A,2 + B,2 + B + C,D) [-1 + C >= 0 && -2 + B + C >= 0 && -1*B + C >= 0 && -1 + A + C >= 0 && -1 + -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= C && B >= 0] f(A,B,C,D) -> end(A,B,C,D) [-1 + C >= 0 && -2 + B + C >= 0 && -1*B + C >= 0 && -1 + A + C >= 0 && -1 + -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1 + D] Signature: {(end,4);(f,4);(sqrt,4)} Rule Graph: [0->{1,2},1->{1,2},2->{}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2] | `- p:[1] c: [1] * Step 3: CloseWith YES + Considered Problem: (Rules: sqrt(A,B,C,D) -> f(0,1,1,D) True f(A,B,C,D) -> f(1 + A,2 + B,2 + B + C,D) [-1 + C >= 0 && -2 + B + C >= 0 && -1*B + C >= 0 && -1 + A + C >= 0 && -1 + -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= C && B >= 0] f(A,B,C,D) -> end(A,B,C,D) [-1 + C >= 0 && -2 + B + C >= 0 && -1*B + C >= 0 && -1 + A + C >= 0 && -1 + -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1 + D] Signature: {(end,4);(f,4);(sqrt,4)} Rule Graph: [0->{1,2},1->{1,2},2->{}] ,We construct a looptree: P: [0,1,2] | `- p:[1] c: [1]) + Applied Processor: CloseWith True + Details: () YES