YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. evalNestedSinglestart(A,B,C) -> evalNestedSingleentryin(A,B,C) True (1,1) 1. evalNestedSingleentryin(A,B,C) -> evalNestedSinglebb5in(0,B,C) True (?,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [A >= 0 && B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (?,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (?,1) Signature: {(evalNestedSinglebb1in,3) ;(evalNestedSinglebb2in,3) ;(evalNestedSinglebb3in,3) ;(evalNestedSinglebb4in,3) ;(evalNestedSinglebb5in,3) ;(evalNestedSingleentryin,3) ;(evalNestedSinglereturnin,3) ;(evalNestedSinglestart,3) ;(evalNestedSinglestop,3)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. evalNestedSinglestart(A,B,C) -> evalNestedSingleentryin(A,B,C) True (1,1) 1. evalNestedSingleentryin(A,B,C) -> evalNestedSinglebb5in(0,B,C) True (1,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [A >= 0 && B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (1,1) Signature: {(evalNestedSinglebb1in,3) ;(evalNestedSinglebb2in,3) ;(evalNestedSinglebb3in,3) ;(evalNestedSinglebb4in,3) ;(evalNestedSinglebb5in,3) ;(evalNestedSingleentryin,3) ;(evalNestedSinglereturnin,3) ;(evalNestedSinglestart,3) ;(evalNestedSinglestop,3)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(2,4)] * Step 3: Looptree YES + Considered Problem: Rules: 0. evalNestedSinglestart(A,B,C) -> evalNestedSingleentryin(A,B,C) True (1,1) 1. evalNestedSingleentryin(A,B,C) -> evalNestedSinglebb5in(0,B,C) True (1,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [A >= 0 && B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (1,1) Signature: {(evalNestedSinglebb1in,3) ;(evalNestedSinglebb2in,3) ;(evalNestedSinglebb3in,3) ;(evalNestedSinglebb4in,3) ;(evalNestedSinglebb5in,3) ;(evalNestedSingleentryin,3) ;(evalNestedSinglereturnin,3) ;(evalNestedSinglestart,3) ;(evalNestedSinglestop,3)} Flow Graph: [0->{1},1->{2,3},2->{5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4,5,6,7,8,9,10,11] | `- p:[2,10,4,9,6,5,7,8] c: [10] | `- p:[5,9,6,7] c: [9] YES