YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb3in(B,A,0) [A >= 1 && B >= 1] (?,1) 2. evalfbb3in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] (?,1) 3. evalfbb3in(A,B,C) -> evalfbb4in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] (?,1) 4. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] 5. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] (?,1) 6. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] (?,1) 7. evalfbbin(A,B,C) -> evalfbb1in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] 8. evalfbbin(A,B,C) -> evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] (?,1) 9. evalfbb1in(A,B,C) -> evalfbb3in(A,-1 + B,1 + C) [-1 + A + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] 10. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] (?,1) Signature: {(evalfbb1in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfbbin,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{10},3->{4,5,6},4->{7,8},5->{7,8},6->{10},7->{9},8->{2,3},9->{2,3},10->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb3in(B,A,0) [A >= 1 && B >= 1] (1,1) 2. evalfbb3in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] (1,1) 3. evalfbb3in(A,B,C) -> evalfbb4in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] (?,1) 4. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] 5. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] (?,1) 6. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] (1,1) 7. evalfbbin(A,B,C) -> evalfbb1in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] 8. evalfbbin(A,B,C) -> evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] (?,1) 9. evalfbb1in(A,B,C) -> evalfbb3in(A,-1 + B,1 + C) [-1 + A + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] 10. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfbbin,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{10},3->{4,5,6},4->{7,8},5->{7,8},6->{10},7->{9},8->{2,3},9->{2,3},10->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(1,2),(8,2)] * Step 3: Looptree YES + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb3in(B,A,0) [A >= 1 && B >= 1] (1,1) 2. evalfbb3in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] (1,1) 3. evalfbb3in(A,B,C) -> evalfbb4in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] (?,1) 4. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] 5. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] (?,1) 6. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] (1,1) 7. evalfbbin(A,B,C) -> evalfbb1in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] 8. evalfbbin(A,B,C) -> evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] (?,1) 9. evalfbb1in(A,B,C) -> evalfbb3in(A,-1 + B,1 + C) [-1 + A + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] 10. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfbbin,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{3},2->{10},3->{4,5,6},4->{7,8},5->{7,8},6->{10},7->{9},8->{3},9->{2,3},10->{}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4,5,6,7,8,9,10] | `- p:[3,8,4,5,9,7] c: [9] | `- p:[3,8,4,5] c: [8] YES