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) -> evalfbb5in(B,B,C,D) True (?,1) 2. evalfbb5in(A,B,C,D) -> evalfbbin(A,B,C,D) [B >= 2] (?,1) 3. evalfbb5in(A,B,C,D) -> evalfreturnin(A,B,C,D) [1 >= B] (?,1) 4. evalfbbin(A,B,C,D) -> evalfbb2in(A,B,-1 + B,-1 + A + B) [-2 + B >= 0] (?,1) 5. evalfbb2in(A,B,C,D) -> evalfbb4in(A,B,C,D) [-1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0 && C >= 1 + D] (?,1) 6. evalfbb2in(A,B,C,D) -> evalfbb3in(A,B,C,D) [-1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0 && D >= C] (?,1) 7. evalfbb3in(A,B,C,D) -> evalfbb1in(A,B,C,D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -3 + B + D >= 0 && 1 + -1*B + D >= 0 && -1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0 && 0 >= 1 + E] 8. evalfbb3in(A,B,C,D) -> evalfbb1in(A,B,C,D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -3 + B + D >= 0 && 1 + -1*B + D >= 0 && -1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0 && E >= 1] 9. evalfbb3in(A,B,C,D) -> evalfbb4in(A,B,C,D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -3 + B + D >= 0 && 1 + -1*B + D >= 0 && -1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0] 10. evalfbb1in(A,B,C,D) -> evalfbb2in(A,B,C,-1 + D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -3 + B + D >= 0 && 1 + -1*B + D >= 0 && -1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0] 11. evalfbb4in(A,B,C,D) -> evalfbb5in(1 + -1*C + D,-1 + C,C,D) [-1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0] (?,1) 12. evalfreturnin(A,B,C,D) -> evalfstop(A,B,C,D) [1 + -1*B >= 0] (?,1) Signature: {(evalfbb1in,4) ;(evalfbb2in,4) ;(evalfbb3in,4) ;(evalfbb4in,4) ;(evalfbb5in,4) ;(evalfbbin,4) ;(evalfentryin,4) ;(evalfreturnin,4) ;(evalfstart,4) ;(evalfstop,4)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{12},4->{5,6},5->{11},6->{7,8,9},7->{10},8->{10},9->{11},10->{5,6},11->{2,3} ,12->{}] + 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) -> evalfbb5in(B,B,C,D) True (1,1) 2. evalfbb5in(A,B,C,D) -> evalfbbin(A,B,C,D) [B >= 2] (?,1) 3. evalfbb5in(A,B,C,D) -> evalfreturnin(A,B,C,D) [1 >= B] (1,1) 4. evalfbbin(A,B,C,D) -> evalfbb2in(A,B,-1 + B,-1 + A + B) [-2 + B >= 0] (?,1) 5. evalfbb2in(A,B,C,D) -> evalfbb4in(A,B,C,D) [-1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0 && C >= 1 + D] (?,1) 6. evalfbb2in(A,B,C,D) -> evalfbb3in(A,B,C,D) [-1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0 && D >= C] (?,1) 7. evalfbb3in(A,B,C,D) -> evalfbb1in(A,B,C,D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -3 + B + D >= 0 && 1 + -1*B + D >= 0 && -1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0 && 0 >= 1 + E] 8. evalfbb3in(A,B,C,D) -> evalfbb1in(A,B,C,D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -3 + B + D >= 0 && 1 + -1*B + D >= 0 && -1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0 && E >= 1] 9. evalfbb3in(A,B,C,D) -> evalfbb4in(A,B,C,D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -3 + B + D >= 0 && 1 + -1*B + D >= 0 && -1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0] 10. evalfbb1in(A,B,C,D) -> evalfbb2in(A,B,C,-1 + D) [-1 + D >= 0 (?,1) && -2 + C + D >= 0 && -1*C + D >= 0 && -3 + B + D >= 0 && 1 + -1*B + D >= 0 && -1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0] 11. evalfbb4in(A,B,C,D) -> evalfbb5in(1 + -1*C + D,-1 + C,C,D) [-1 + B + -1*C >= 0 && -1 + C >= 0 && -3 + B + C >= 0 && 1 + -1*B + C >= 0 && -2 + B >= 0] (?,1) 12. evalfreturnin(A,B,C,D) -> evalfstop(A,B,C,D) [1 + -1*B >= 0] (1,1) Signature: {(evalfbb1in,4) ;(evalfbb2in,4) ;(evalfbb3in,4) ;(evalfbb4in,4) ;(evalfbb5in,4) ;(evalfbbin,4) ;(evalfentryin,4) ;(evalfreturnin,4) ;(evalfstart,4) ;(evalfstop,4)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{12},4->{5,6},5->{11},6->{7,8,9},7->{10},8->{10},9->{11},10->{5,6},11->{2,3} ,12->{}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4,5,6,7,8,9,10,11,12] | `- p:[2,11,5,4,10,7,6,8,9] c: [11] | `- p:[6,10,7,8] c: [10] YES