MAYBE * Step 1: UnsatPaths MAYBE + Considered Problem: Rules: 0. f5(A,B,C,D,E) -> f300(A,B,C,D,E) True (1,1) 1. f4(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && A >= B] (?,1) 2. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && F >= 1 && B >= 1 + A] (?,1) 3. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] (?,1) 4. f4(A,B,C,D,E) -> f2(A,B,C,0,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && B >= 1 + A] (?,1) 5. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (?,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && F >= 1 && B >= 1 + A] 6. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (?,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] 7. f2(A,B,C,D,E) -> f2(A,B,C,0,E) [-1*D >= 0 && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && B >= 1 + A] (?,1) 8. f300(A,B,C,D,E) -> f1(A,B,C,D,F) [C >= A] (?,1) 9. f300(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [A >= 1 + C && A >= B] (?,1) 10. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [F >= 1 && A >= 1 + C && B >= 1 + A] (?,1) 11. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [0 >= 1 + F && A >= 1 + C && B >= 1 + A] (?,1) 12. f300(A,B,C,D,E) -> f2(A,B,C,0,E) [A >= 1 + C && B >= 1 + A] (?,1) Signature: {(f1,5);(f2,5);(f300,5);(f4,5);(f5,5)} Flow Graph: [0->{8,9,10,11,12},1->{8,9,10,11,12},2->{1,2,3,4},3->{1,2,3,4},4->{5,6,7},5->{1,2,3,4},6->{1,2,3,4},7->{5 ,6,7},8->{},9->{8,9,10,11,12},10->{1,2,3,4},11->{1,2,3,4},12->{5,6,7}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(1,8),(1,10),(1,11),(1,12),(9,10),(9,11),(9,12)] * Step 2: TrivialSCCs MAYBE + Considered Problem: Rules: 0. f5(A,B,C,D,E) -> f300(A,B,C,D,E) True (1,1) 1. f4(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && A >= B] (?,1) 2. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && F >= 1 && B >= 1 + A] (?,1) 3. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] (?,1) 4. f4(A,B,C,D,E) -> f2(A,B,C,0,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && B >= 1 + A] (?,1) 5. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (?,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && F >= 1 && B >= 1 + A] 6. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (?,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] 7. f2(A,B,C,D,E) -> f2(A,B,C,0,E) [-1*D >= 0 && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && B >= 1 + A] (?,1) 8. f300(A,B,C,D,E) -> f1(A,B,C,D,F) [C >= A] (?,1) 9. f300(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [A >= 1 + C && A >= B] (?,1) 10. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [F >= 1 && A >= 1 + C && B >= 1 + A] (?,1) 11. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [0 >= 1 + F && A >= 1 + C && B >= 1 + A] (?,1) 12. f300(A,B,C,D,E) -> f2(A,B,C,0,E) [A >= 1 + C && B >= 1 + A] (?,1) Signature: {(f1,5);(f2,5);(f300,5);(f4,5);(f5,5)} Flow Graph: [0->{8,9,10,11,12},1->{9},2->{1,2,3,4},3->{1,2,3,4},4->{5,6,7},5->{1,2,3,4},6->{1,2,3,4},7->{5,6,7},8->{} ,9->{8,9},10->{1,2,3,4},11->{1,2,3,4},12->{5,6,7}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 3: PolyRank MAYBE + Considered Problem: Rules: 0. f5(A,B,C,D,E) -> f300(A,B,C,D,E) True (1,1) 1. f4(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && A >= B] (1,1) 2. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && F >= 1 && B >= 1 + A] (?,1) 3. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] (?,1) 4. f4(A,B,C,D,E) -> f2(A,B,C,0,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && B >= 1 + A] (?,1) 5. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (?,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && F >= 1 && B >= 1 + A] 6. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (?,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] 7. f2(A,B,C,D,E) -> f2(A,B,C,0,E) [-1*D >= 0 && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && B >= 1 + A] (?,1) 8. f300(A,B,C,D,E) -> f1(A,B,C,D,F) [C >= A] (1,1) 9. f300(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [A >= 1 + C && A >= B] (?,1) 10. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [F >= 1 && A >= 1 + C && B >= 1 + A] (1,1) 11. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [0 >= 1 + F && A >= 1 + C && B >= 1 + A] (1,1) 12. f300(A,B,C,D,E) -> f2(A,B,C,0,E) [A >= 1 + C && B >= 1 + A] (1,1) Signature: {(f1,5);(f2,5);(f300,5);(f4,5);(f5,5)} Flow Graph: [0->{8,9,10,11,12},1->{9},2->{1,2,3,4},3->{1,2,3,4},4->{5,6,7},5->{1,2,3,4},6->{1,2,3,4},7->{5,6,7},8->{} ,9->{8,9},10->{1,2,3,4},11->{1,2,3,4},12->{5,6,7}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = -2*x1 + 3*x2 + -1*x3 p(f2) = -2*x1 + 3*x2 + -1*x3 p(f300) = -2*x1 + 3*x2 + -1*x3 p(f4) = 1 + -2*x1 + 3*x2 + -1*x3 p(f5) = -2*x1 + 3*x2 + -1*x3 Following rules are strictly oriented: [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && A >= B] ==> f4(A,B,C,D,E) = 1 + -2*A + 3*B + -1*C > -1 + -2*A + 3*B + -1*C = f300(A,B,1 + C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] ==> f4(A,B,C,D,E) = 1 + -2*A + 3*B + -1*C > -1 + -2*A + 3*B + -1*C = f4(1 + A,B,C,F,E) [-1*D >= 0 ==> && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && F >= 1 && B >= 1 + A] f2(A,B,C,D,E) = -2*A + 3*B + -1*C > -1 + -2*A + 3*B + -1*C = f4(1 + A,B,C,F,E) Following rules are weakly oriented: True ==> f5(A,B,C,D,E) = -2*A + 3*B + -1*C >= -2*A + 3*B + -1*C = f300(A,B,C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && F >= 1 && B >= 1 + A] ==> f4(A,B,C,D,E) = 1 + -2*A + 3*B + -1*C >= -1 + -2*A + 3*B + -1*C = f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && B >= 1 + A] ==> f4(A,B,C,D,E) = 1 + -2*A + 3*B + -1*C >= -2*A + 3*B + -1*C = f2(A,B,C,0,E) [-1*D >= 0 ==> && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] f2(A,B,C,D,E) = -2*A + 3*B + -1*C >= -1 + -2*A + 3*B + -1*C = f4(1 + A,B,C,F,E) [-1*D >= 0 && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && B >= 1 + A] ==> f2(A,B,C,D,E) = -2*A + 3*B + -1*C >= -2*A + 3*B + -1*C = f2(A,B,C,0,E) [C >= A] ==> f300(A,B,C,D,E) = -2*A + 3*B + -1*C >= -2*A + 3*B + -1*C = f1(A,B,C,D,F) [A >= 1 + C && A >= B] ==> f300(A,B,C,D,E) = -2*A + 3*B + -1*C >= -1 + -2*A + 3*B + -1*C = f300(A,B,1 + C,D,E) [F >= 1 && A >= 1 + C && B >= 1 + A] ==> f300(A,B,C,D,E) = -2*A + 3*B + -1*C >= -1 + -2*A + 3*B + -1*C = f4(1 + A,B,C,F,E) [0 >= 1 + F && A >= 1 + C && B >= 1 + A] ==> f300(A,B,C,D,E) = -2*A + 3*B + -1*C >= -1 + -2*A + 3*B + -1*C = f4(1 + A,B,C,F,E) [A >= 1 + C && B >= 1 + A] ==> f300(A,B,C,D,E) = -2*A + 3*B + -1*C >= -2*A + 3*B + -1*C = f2(A,B,C,0,E) * Step 4: PolyRank MAYBE + Considered Problem: Rules: 0. f5(A,B,C,D,E) -> f300(A,B,C,D,E) True (1,1) 1. f4(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && A >= B] (1,1) 2. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && F >= 1 && B >= 1 + A] (?,1) 3. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] (2*A + 3*B + C,1) 4. f4(A,B,C,D,E) -> f2(A,B,C,0,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && B >= 1 + A] (?,1) 5. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (2*A + 3*B + C,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && F >= 1 && B >= 1 + A] 6. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (?,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] 7. f2(A,B,C,D,E) -> f2(A,B,C,0,E) [-1*D >= 0 && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && B >= 1 + A] (?,1) 8. f300(A,B,C,D,E) -> f1(A,B,C,D,F) [C >= A] (1,1) 9. f300(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [A >= 1 + C && A >= B] (?,1) 10. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [F >= 1 && A >= 1 + C && B >= 1 + A] (1,1) 11. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [0 >= 1 + F && A >= 1 + C && B >= 1 + A] (1,1) 12. f300(A,B,C,D,E) -> f2(A,B,C,0,E) [A >= 1 + C && B >= 1 + A] (1,1) Signature: {(f1,5);(f2,5);(f300,5);(f4,5);(f5,5)} Flow Graph: [0->{8,9,10,11,12},1->{9},2->{1,2,3,4},3->{1,2,3,4},4->{5,6,7},5->{1,2,3,4},6->{1,2,3,4},7->{5,6,7},8->{} ,9->{8,9},10->{1,2,3,4},11->{1,2,3,4},12->{5,6,7}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = -1*x1 + 3*x2 + -2*x3 p(f2) = -1 + -1*x1 + 3*x2 + -2*x3 p(f300) = -1*x1 + 3*x2 + -2*x3 p(f4) = -1*x1 + 3*x2 + -2*x3 p(f5) = -1*x1 + 3*x2 + -2*x3 Following rules are strictly oriented: [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && A >= B] ==> f4(A,B,C,D,E) = -1*A + 3*B + -2*C > -2 + -1*A + 3*B + -2*C = f300(A,B,1 + C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && F >= 1 && B >= 1 + A] ==> f4(A,B,C,D,E) = -1*A + 3*B + -2*C > -1 + -1*A + 3*B + -2*C = f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] ==> f4(A,B,C,D,E) = -1*A + 3*B + -2*C > -1 + -1*A + 3*B + -2*C = f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && B >= 1 + A] ==> f4(A,B,C,D,E) = -1*A + 3*B + -2*C > -1 + -1*A + 3*B + -2*C = f2(A,B,C,0,E) Following rules are weakly oriented: True ==> f5(A,B,C,D,E) = -1*A + 3*B + -2*C >= -1*A + 3*B + -2*C = f300(A,B,C,D,E) [-1*D >= 0 ==> && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && F >= 1 && B >= 1 + A] f2(A,B,C,D,E) = -1 + -1*A + 3*B + -2*C >= -1 + -1*A + 3*B + -2*C = f4(1 + A,B,C,F,E) [-1*D >= 0 ==> && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] f2(A,B,C,D,E) = -1 + -1*A + 3*B + -2*C >= -1 + -1*A + 3*B + -2*C = f4(1 + A,B,C,F,E) [-1*D >= 0 && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && B >= 1 + A] ==> f2(A,B,C,D,E) = -1 + -1*A + 3*B + -2*C >= -1 + -1*A + 3*B + -2*C = f2(A,B,C,0,E) [C >= A] ==> f300(A,B,C,D,E) = -1*A + 3*B + -2*C >= -1*A + 3*B + -2*C = f1(A,B,C,D,F) [A >= 1 + C && A >= B] ==> f300(A,B,C,D,E) = -1*A + 3*B + -2*C >= -2 + -1*A + 3*B + -2*C = f300(A,B,1 + C,D,E) [F >= 1 && A >= 1 + C && B >= 1 + A] ==> f300(A,B,C,D,E) = -1*A + 3*B + -2*C >= -1 + -1*A + 3*B + -2*C = f4(1 + A,B,C,F,E) [0 >= 1 + F && A >= 1 + C && B >= 1 + A] ==> f300(A,B,C,D,E) = -1*A + 3*B + -2*C >= -1 + -1*A + 3*B + -2*C = f4(1 + A,B,C,F,E) [A >= 1 + C && B >= 1 + A] ==> f300(A,B,C,D,E) = -1*A + 3*B + -2*C >= -1 + -1*A + 3*B + -2*C = f2(A,B,C,0,E) * Step 5: PolyRank MAYBE + Considered Problem: Rules: 0. f5(A,B,C,D,E) -> f300(A,B,C,D,E) True (1,1) 1. f4(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && A >= B] (1,1) 2. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && F >= 1 && B >= 1 + A] (A + 3*B + 2*C,1) 3. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] (2*A + 3*B + C,1) 4. f4(A,B,C,D,E) -> f2(A,B,C,0,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && B >= 1 + A] (A + 3*B + 2*C,1) 5. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (2*A + 3*B + C,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && F >= 1 && B >= 1 + A] 6. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (?,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] 7. f2(A,B,C,D,E) -> f2(A,B,C,0,E) [-1*D >= 0 && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && B >= 1 + A] (?,1) 8. f300(A,B,C,D,E) -> f1(A,B,C,D,F) [C >= A] (1,1) 9. f300(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [A >= 1 + C && A >= B] (?,1) 10. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [F >= 1 && A >= 1 + C && B >= 1 + A] (1,1) 11. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [0 >= 1 + F && A >= 1 + C && B >= 1 + A] (1,1) 12. f300(A,B,C,D,E) -> f2(A,B,C,0,E) [A >= 1 + C && B >= 1 + A] (1,1) Signature: {(f1,5);(f2,5);(f300,5);(f4,5);(f5,5)} Flow Graph: [0->{8,9,10,11,12},1->{9},2->{1,2,3,4},3->{1,2,3,4},4->{5,6,7},5->{1,2,3,4},6->{1,2,3,4},7->{5,6,7},8->{} ,9->{8,9},10->{1,2,3,4},11->{1,2,3,4},12->{5,6,7}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = -1*x1 + 2*x2 + -1*x3 p(f2) = -1 + -1*x1 + 2*x2 + -1*x3 p(f300) = -1*x1 + 2*x2 + -1*x3 p(f4) = -1 + -1*x1 + 2*x2 + -1*x3 p(f5) = -1*x1 + 2*x2 + -1*x3 Following rules are strictly oriented: [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] ==> f4(A,B,C,D,E) = -1 + -1*A + 2*B + -1*C > -2 + -1*A + 2*B + -1*C = f4(1 + A,B,C,F,E) [-1*D >= 0 ==> && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && F >= 1 && B >= 1 + A] f2(A,B,C,D,E) = -1 + -1*A + 2*B + -1*C > -2 + -1*A + 2*B + -1*C = f4(1 + A,B,C,F,E) [-1*D >= 0 ==> && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] f2(A,B,C,D,E) = -1 + -1*A + 2*B + -1*C > -2 + -1*A + 2*B + -1*C = f4(1 + A,B,C,F,E) [F >= 1 && A >= 1 + C && B >= 1 + A] ==> f300(A,B,C,D,E) = -1*A + 2*B + -1*C > -2 + -1*A + 2*B + -1*C = f4(1 + A,B,C,F,E) [0 >= 1 + F && A >= 1 + C && B >= 1 + A] ==> f300(A,B,C,D,E) = -1*A + 2*B + -1*C > -2 + -1*A + 2*B + -1*C = f4(1 + A,B,C,F,E) [A >= 1 + C && B >= 1 + A] ==> f300(A,B,C,D,E) = -1*A + 2*B + -1*C > -1 + -1*A + 2*B + -1*C = f2(A,B,C,0,E) Following rules are weakly oriented: True ==> f5(A,B,C,D,E) = -1*A + 2*B + -1*C >= -1*A + 2*B + -1*C = f300(A,B,C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && A >= B] ==> f4(A,B,C,D,E) = -1 + -1*A + 2*B + -1*C >= -1 + -1*A + 2*B + -1*C = f300(A,B,1 + C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && F >= 1 && B >= 1 + A] ==> f4(A,B,C,D,E) = -1 + -1*A + 2*B + -1*C >= -2 + -1*A + 2*B + -1*C = f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && B >= 1 + A] ==> f4(A,B,C,D,E) = -1 + -1*A + 2*B + -1*C >= -1 + -1*A + 2*B + -1*C = f2(A,B,C,0,E) [-1*D >= 0 && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && B >= 1 + A] ==> f2(A,B,C,D,E) = -1 + -1*A + 2*B + -1*C >= -1 + -1*A + 2*B + -1*C = f2(A,B,C,0,E) [C >= A] ==> f300(A,B,C,D,E) = -1*A + 2*B + -1*C >= -1*A + 2*B + -1*C = f1(A,B,C,D,F) [A >= 1 + C && A >= B] ==> f300(A,B,C,D,E) = -1*A + 2*B + -1*C >= -1 + -1*A + 2*B + -1*C = f300(A,B,1 + C,D,E) * Step 6: Failure MAYBE + Considered Problem: Rules: 0. f5(A,B,C,D,E) -> f300(A,B,C,D,E) True (1,1) 1. f4(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && A >= B] (1,1) 2. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && F >= 1 && B >= 1 + A] (A + 3*B + 2*C,1) 3. f4(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] (2*A + 3*B + C,1) 4. f4(A,B,C,D,E) -> f2(A,B,C,0,E) [-2 + B + -1*C >= 0 && -2 + A + -1*C >= 0 && -1*A + B >= 0 && B >= 1 + A] (A + 3*B + 2*C,1) 5. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (2*A + 3*B + C,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && F >= 1 && B >= 1 + A] 6. f2(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [-1*D >= 0 (A + 2*B + C,1) && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && 0 >= 1 + F && B >= 1 + A] 7. f2(A,B,C,D,E) -> f2(A,B,C,0,E) [-1*D >= 0 && D >= 0 && -2 + B + -1*C >= 0 && -1 + A + -1*C >= 0 && -1 + -1*A + B >= 0 && B >= 1 + A] (?,1) 8. f300(A,B,C,D,E) -> f1(A,B,C,D,F) [C >= A] (1,1) 9. f300(A,B,C,D,E) -> f300(A,B,1 + C,D,E) [A >= 1 + C && A >= B] (?,1) 10. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [F >= 1 && A >= 1 + C && B >= 1 + A] (1,1) 11. f300(A,B,C,D,E) -> f4(1 + A,B,C,F,E) [0 >= 1 + F && A >= 1 + C && B >= 1 + A] (1,1) 12. f300(A,B,C,D,E) -> f2(A,B,C,0,E) [A >= 1 + C && B >= 1 + A] (1,1) Signature: {(f1,5);(f2,5);(f300,5);(f4,5);(f5,5)} Flow Graph: [0->{8,9,10,11,12},1->{9},2->{1,2,3,4},3->{1,2,3,4},4->{5,6,7},5->{1,2,3,4},6->{1,2,3,4},7->{5,6,7},8->{} ,9->{8,9},10->{1,2,3,4},11->{1,2,3,4},12->{5,6,7}] + Applied Processor: Failing "Open problems left." + Details: Open problems left. MAYBE