YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. evalEx4start(A,B,C,D) -> evalEx4entryin(A,B,C,D) True (1,1) 1. evalEx4entryin(A,B,C,D) -> evalEx4bb4in(1,A,C,D) True (?,1) 2. evalEx4bb4in(A,B,C,D) -> evalEx4bb2in(A,B,0,B) [A = 1] (?,1) 3. evalEx4bb4in(A,B,C,D) -> evalEx4returnin(A,B,C,D) [0 >= A] (?,1) 4. evalEx4bb4in(A,B,C,D) -> evalEx4returnin(A,B,C,D) [A >= 2] (?,1) 5. evalEx4bb2in(A,B,C,D) -> evalEx4bb4in(C,D,C,D) [B + -1*D >= 0 && C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && 0 >= D] (?,1) 6. evalEx4bb2in(A,B,C,D) -> evalEx4bb3in(A,B,C,D) [B + -1*D >= 0 && C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && D >= 1] (?,1) 7. evalEx4bb3in(A,B,C,D) -> evalEx4bb1in(A,B,C,D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && 0 >= 1 + E] 8. evalEx4bb3in(A,B,C,D) -> evalEx4bb1in(A,B,C,D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && E >= 1] 9. evalEx4bb3in(A,B,C,D) -> evalEx4bb4in(C,D,C,D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0] 10. evalEx4bb1in(A,B,C,D) -> evalEx4bb2in(A,B,1,-1 + D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0] 11. evalEx4returnin(A,B,C,D) -> evalEx4stop(A,B,C,D) True (?,1) Signature: {(evalEx4bb1in,4) ;(evalEx4bb2in,4) ;(evalEx4bb3in,4) ;(evalEx4bb4in,4) ;(evalEx4entryin,4) ;(evalEx4returnin,4) ;(evalEx4start,4) ;(evalEx4stop,4)} Flow Graph: [0->{1},1->{2,3,4},2->{5,6},3->{11},4->{11},5->{2,3,4},6->{7,8,9},7->{10},8->{10},9->{2,3,4},10->{5,6} ,11->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. evalEx4start(A,B,C,D) -> evalEx4entryin(A,B,C,D) True (1,1) 1. evalEx4entryin(A,B,C,D) -> evalEx4bb4in(1,A,C,D) True (1,1) 2. evalEx4bb4in(A,B,C,D) -> evalEx4bb2in(A,B,0,B) [A = 1] (?,1) 3. evalEx4bb4in(A,B,C,D) -> evalEx4returnin(A,B,C,D) [0 >= A] (1,1) 4. evalEx4bb4in(A,B,C,D) -> evalEx4returnin(A,B,C,D) [A >= 2] (1,1) 5. evalEx4bb2in(A,B,C,D) -> evalEx4bb4in(C,D,C,D) [B + -1*D >= 0 && C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && 0 >= D] (?,1) 6. evalEx4bb2in(A,B,C,D) -> evalEx4bb3in(A,B,C,D) [B + -1*D >= 0 && C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && D >= 1] (?,1) 7. evalEx4bb3in(A,B,C,D) -> evalEx4bb1in(A,B,C,D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && 0 >= 1 + E] 8. evalEx4bb3in(A,B,C,D) -> evalEx4bb1in(A,B,C,D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && E >= 1] 9. evalEx4bb3in(A,B,C,D) -> evalEx4bb4in(C,D,C,D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0] 10. evalEx4bb1in(A,B,C,D) -> evalEx4bb2in(A,B,1,-1 + D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0] 11. evalEx4returnin(A,B,C,D) -> evalEx4stop(A,B,C,D) True (1,1) Signature: {(evalEx4bb1in,4) ;(evalEx4bb2in,4) ;(evalEx4bb3in,4) ;(evalEx4bb4in,4) ;(evalEx4entryin,4) ;(evalEx4returnin,4) ;(evalEx4start,4) ;(evalEx4stop,4)} Flow Graph: [0->{1},1->{2,3,4},2->{5,6},3->{11},4->{11},5->{2,3,4},6->{7,8,9},7->{10},8->{10},9->{2,3,4},10->{5,6} ,11->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(1,3),(1,4)] * Step 3: Looptree YES + Considered Problem: Rules: 0. evalEx4start(A,B,C,D) -> evalEx4entryin(A,B,C,D) True (1,1) 1. evalEx4entryin(A,B,C,D) -> evalEx4bb4in(1,A,C,D) True (1,1) 2. evalEx4bb4in(A,B,C,D) -> evalEx4bb2in(A,B,0,B) [A = 1] (?,1) 3. evalEx4bb4in(A,B,C,D) -> evalEx4returnin(A,B,C,D) [0 >= A] (1,1) 4. evalEx4bb4in(A,B,C,D) -> evalEx4returnin(A,B,C,D) [A >= 2] (1,1) 5. evalEx4bb2in(A,B,C,D) -> evalEx4bb4in(C,D,C,D) [B + -1*D >= 0 && C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && 0 >= D] (?,1) 6. evalEx4bb2in(A,B,C,D) -> evalEx4bb3in(A,B,C,D) [B + -1*D >= 0 && C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && D >= 1] (?,1) 7. evalEx4bb3in(A,B,C,D) -> evalEx4bb1in(A,B,C,D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && 0 >= 1 + E] 8. evalEx4bb3in(A,B,C,D) -> evalEx4bb1in(A,B,C,D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0 && E >= 1] 9. evalEx4bb3in(A,B,C,D) -> evalEx4bb4in(C,D,C,D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0] 10. evalEx4bb1in(A,B,C,D) -> evalEx4bb2in(A,B,1,-1 + D) [B + -1*D >= 0 (?,1) && -1 + D >= 0 && -1 + C + D >= 0 && -2 + B + D >= 0 && -2 + A + D >= 0 && -1*A + D >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && 1 + -1*A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1*A + B >= 0 && 1 + -1*A >= 0 && -1 + A >= 0] 11. evalEx4returnin(A,B,C,D) -> evalEx4stop(A,B,C,D) True (1,1) Signature: {(evalEx4bb1in,4) ;(evalEx4bb2in,4) ;(evalEx4bb3in,4) ;(evalEx4bb4in,4) ;(evalEx4entryin,4) ;(evalEx4returnin,4) ;(evalEx4start,4) ;(evalEx4stop,4)} Flow Graph: [0->{1},1->{2},2->{5,6},3->{11},4->{11},5->{2,3,4},6->{7,8,9},7->{10},8->{10},9->{2,3,4},10->{5,6},11->{}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4,5,6,7,8,9,10,11] | `- p:[2,5,10,7,6,8,9] c: [10] | `- p:[2,5,9,6] c: [9] | `- p:[2,5] c: [5] YES