YES(?,O(1)) * Step 1: UnsatPaths WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A) -> f3(0) True (1,1) 1. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A] (?,1) 2. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A && 0 >= 1 + B] (?,1) 3. f3(A) -> f13(A) [A >= 0 && A >= 42] (?,1) Signature: {(f0,1);(f13,1);(f3,1)} Flow Graph: [0->{1,2,3},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,3)] * Step 2: TrivialSCCs WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A) -> f3(0) True (1,1) 1. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A] (?,1) 2. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A && 0 >= 1 + B] (?,1) 3. f3(A) -> f13(A) [A >= 0 && A >= 42] (?,1) Signature: {(f0,1);(f13,1);(f3,1)} Flow Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 3: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A) -> f3(0) True (1,1) 1. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A] (?,1) 2. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A && 0 >= 1 + B] (?,1) 3. f3(A) -> f13(A) [A >= 0 && A >= 42] (1,1) Signature: {(f0,1);(f13,1);(f3,1)} Flow Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = 1723 p(f13) = 1723 + -42*x1 p(f3) = 1723 + -42*x1 Following rules are strictly oriented: [A >= 0 && 41 >= A && 0 >= 1 + B] ==> f3(A) = 1723 + -42*A > 1681 + -42*A = f3(1 + A) Following rules are weakly oriented: True ==> f0(A) = 1723 >= 1723 = f3(0) [A >= 0 && 41 >= A] ==> f3(A) = 1723 + -42*A >= 1681 + -42*A = f3(1 + A) [A >= 0 && A >= 42] ==> f3(A) = 1723 + -42*A >= 1723 + -42*A = f13(A) * Step 4: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A) -> f3(0) True (1,1) 1. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A] (?,1) 2. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A && 0 >= 1 + B] (1723,1) 3. f3(A) -> f13(A) [A >= 0 && A >= 42] (1,1) Signature: {(f0,1);(f13,1);(f3,1)} Flow Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = 1723 p(f13) = 1723 + -42*x1 p(f3) = 1723 + -42*x1 Following rules are strictly oriented: [A >= 0 && 41 >= A] ==> f3(A) = 1723 + -42*A > 1681 + -42*A = f3(1 + A) Following rules are weakly oriented: True ==> f0(A) = 1723 >= 1723 = f3(0) [A >= 0 && 41 >= A && 0 >= 1 + B] ==> f3(A) = 1723 + -42*A >= 1681 + -42*A = f3(1 + A) [A >= 0 && A >= 42] ==> f3(A) = 1723 + -42*A >= 1723 + -42*A = f13(A) * Step 5: KnowledgePropagation WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A) -> f3(0) True (1,1) 1. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A] (1723,1) 2. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A && 0 >= 1 + B] (1723,1) 3. f3(A) -> f13(A) [A >= 0 && A >= 42] (1,1) Signature: {(f0,1);(f13,1);(f3,1)} Flow Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(1))