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) -> evalfbb4in(1,B,C) True (?,1) 2. evalfbb4in(A,B,C) -> evalfbb2in(A,B,A) [-1 + A >= 0 && B >= A] (?,1) 3. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [-1 + A >= 0 && A >= 1 + B] (?,1) 4. evalfbb2in(A,B,C) -> evalfbb1in(A,B,C) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && B >= C] 5. evalfbb2in(A,B,C) -> evalfbb3in(A,B,C) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && C >= 1 + B] 6. evalfbb1in(A,B,C) -> evalfbb2in(A,B,1 + C) [B + -1*C >= 0 (?,1) && -1 + C >= 0 && -2 + B + C >= 0 && -2 + A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 7. evalfbb3in(A,B,C) -> evalfbb4in(1 + A,B,C) [-2 + C >= 0 (?,1) && -3 + B + C >= 0 && -1 + -1*B + C >= 0 && -3 + A + C >= 0 && -1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 8. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [-1 + A + -1*B >= 0 && -1 + A >= 0] (?,1) Signature: {(evalfbb1in,3) ;(evalfbb2in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} 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: UnsatPaths YES + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb4in(1,B,C) True (1,1) 2. evalfbb4in(A,B,C) -> evalfbb2in(A,B,A) [-1 + A >= 0 && B >= A] (?,1) 3. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [-1 + A >= 0 && A >= 1 + B] (1,1) 4. evalfbb2in(A,B,C) -> evalfbb1in(A,B,C) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && B >= C] 5. evalfbb2in(A,B,C) -> evalfbb3in(A,B,C) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && C >= 1 + B] 6. evalfbb1in(A,B,C) -> evalfbb2in(A,B,1 + C) [B + -1*C >= 0 (?,1) && -1 + C >= 0 && -2 + B + C >= 0 && -2 + A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 7. evalfbb3in(A,B,C) -> evalfbb4in(1 + A,B,C) [-2 + C >= 0 (?,1) && -3 + B + C >= 0 && -1 + -1*B + C >= 0 && -3 + A + C >= 0 && -1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 8. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [-1 + A + -1*B >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb2in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} 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: UnsatPaths + Details: We remove following edges from the transition graph: [(2,5)] * Step 3: Looptree YES + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb4in(1,B,C) True (1,1) 2. evalfbb4in(A,B,C) -> evalfbb2in(A,B,A) [-1 + A >= 0 && B >= A] (?,1) 3. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [-1 + A >= 0 && A >= 1 + B] (1,1) 4. evalfbb2in(A,B,C) -> evalfbb1in(A,B,C) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && B >= C] 5. evalfbb2in(A,B,C) -> evalfbb3in(A,B,C) [-1 + C >= 0 (?,1) && -2 + B + C >= 0 && -2 + A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0 && C >= 1 + B] 6. evalfbb1in(A,B,C) -> evalfbb2in(A,B,1 + C) [B + -1*C >= 0 (?,1) && -1 + C >= 0 && -2 + B + C >= 0 && -2 + A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 7. evalfbb3in(A,B,C) -> evalfbb4in(1 + A,B,C) [-2 + C >= 0 (?,1) && -3 + B + C >= 0 && -1 + -1*B + C >= 0 && -3 + A + C >= 0 && -1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && -1 + A >= 0] 8. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [-1 + A + -1*B >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb2in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{4},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