YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. evalfstart(A,B,C,D) -> evalfentryin(A,B,C,D) True (1,1) 1. evalfentryin(A,B,C,D) -> evalfbb4in(1,B,C,D) True (?,1) 2. evalfbb4in(A,B,C,D) -> evalfbb2in(A,B,1,D) [-1 + A >= 0 && B >= A] (?,1) 3. evalfbb4in(A,B,C,D) -> evalfreturnin(A,B,C,D) [-1 + A >= 0 && A >= 1 + B] (?,1) 4. evalfbb2in(A,B,C,D) -> evalfbb1in(A,B,C,D) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && D >= C] 5. evalfbb2in(A,B,C,D) -> evalfbb3in(A,B,C,D) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && C >= 1 + D] 6. evalfbb1in(A,B,C,D) -> evalfbb2in(A,B,1 + C,D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1 + C >= 0 && -2 + B + C >= 0 && -2 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 7. evalfbb3in(A,B,C,D) -> evalfbb4in(1 + A,B,C,D) [-1 + C + -1*D >= 0 (?,1) && -1 + C >= 0 && -2 + B + C >= 0 && -2 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 8. evalfreturnin(A,B,C,D) -> evalfstop(A,B,C,D) [-1 + A + -1*B >= 0 && -1 + A >= 0] (?,1) Signature: {(evalfbb1in,4) ;(evalfbb2in,4) ;(evalfbb3in,4) ;(evalfbb4in,4) ;(evalfentryin,4) ;(evalfreturnin,4) ;(evalfstart,4) ;(evalfstop,4)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{4,5},7->{2,3},8->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: Looptree YES + Considered Problem: Rules: 0. evalfstart(A,B,C,D) -> evalfentryin(A,B,C,D) True (1,1) 1. evalfentryin(A,B,C,D) -> evalfbb4in(1,B,C,D) True (1,1) 2. evalfbb4in(A,B,C,D) -> evalfbb2in(A,B,1,D) [-1 + A >= 0 && B >= A] (?,1) 3. evalfbb4in(A,B,C,D) -> evalfreturnin(A,B,C,D) [-1 + A >= 0 && A >= 1 + B] (1,1) 4. evalfbb2in(A,B,C,D) -> evalfbb1in(A,B,C,D) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && D >= C] 5. evalfbb2in(A,B,C,D) -> evalfbb3in(A,B,C,D) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && C >= 1 + D] 6. evalfbb1in(A,B,C,D) -> evalfbb2in(A,B,1 + C,D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1 + C >= 0 && -2 + B + C >= 0 && -2 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 7. evalfbb3in(A,B,C,D) -> evalfbb4in(1 + A,B,C,D) [-1 + C + -1*D >= 0 (?,1) && -1 + C >= 0 && -2 + B + C >= 0 && -2 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 8. evalfreturnin(A,B,C,D) -> evalfstop(A,B,C,D) [-1 + A + -1*B >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,4) ;(evalfbb2in,4) ;(evalfbb3in,4) ;(evalfbb4in,4) ;(evalfentryin,4) ;(evalfreturnin,4) ;(evalfstart,4) ;(evalfstop,4)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{4,5},7->{2,3},8->{}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4,5,6,7,8] | `- p:[2,7,5,6,4] c: [7] | `- p:[4,6] c: [6] YES