YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f300(A,B,C,D) -> f1(A,B,C,E) [A >= B] (?,1) 1. f300(A,B,C,D) -> f300(A,-1 + B,0,D) [B >= 1 + A] (?,1) 2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1) 3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (?,1) 4. f2(A,B,C,D) -> f300(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f300(A,B,C,D) -> f1(A,B,C,E) [A >= B] (1,1) 1. f300(A,B,C,D) -> f300(A,-1 + B,0,D) [B >= 1 + A] (?,1) 2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1) 3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (?,1) 4. f2(A,B,C,D) -> f300(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = -1*x1 + x2 p(f2) = -1*x1 + x2 p(f300) = -1*x1 + x2 Following rules are strictly oriented: [E >= 1 && B >= 1 + A] ==> f300(A,B,C,D) = -1*A + B > -1 + -1*A + B = f300(1 + A,B,E,D) Following rules are weakly oriented: [A >= B] ==> f300(A,B,C,D) = -1*A + B >= -1*A + B = f1(A,B,C,E) [B >= 1 + A] ==> f300(A,B,C,D) = -1*A + B >= -1 + -1*A + B = f300(A,-1 + B,0,D) [0 >= 1 + E && B >= 1 + A] ==> f300(A,B,C,D) = -1*A + B >= -1 + -1*A + B = f300(1 + A,B,E,D) True ==> f2(A,B,C,D) = -1*A + B >= -1*A + B = f300(A,B,C,D) * Step 3: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f300(A,B,C,D) -> f1(A,B,C,E) [A >= B] (1,1) 1. f300(A,B,C,D) -> f300(A,-1 + B,0,D) [B >= 1 + A] (?,1) 2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1) 3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (A + B,1) 4. f2(A,B,C,D) -> f300(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = -1*x1 + x2 p(f2) = -1*x1 + x2 p(f300) = -1*x1 + x2 Following rules are strictly oriented: [0 >= 1 + E && B >= 1 + A] ==> f300(A,B,C,D) = -1*A + B > -1 + -1*A + B = f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A] ==> f300(A,B,C,D) = -1*A + B > -1 + -1*A + B = f300(1 + A,B,E,D) Following rules are weakly oriented: [A >= B] ==> f300(A,B,C,D) = -1*A + B >= -1*A + B = f1(A,B,C,E) [B >= 1 + A] ==> f300(A,B,C,D) = -1*A + B >= -1 + -1*A + B = f300(A,-1 + B,0,D) True ==> f2(A,B,C,D) = -1*A + B >= -1*A + B = f300(A,B,C,D) * Step 4: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f300(A,B,C,D) -> f1(A,B,C,E) [A >= B] (1,1) 1. f300(A,B,C,D) -> f300(A,-1 + B,0,D) [B >= 1 + A] (?,1) 2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (A + B,1) 3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (A + B,1) 4. f2(A,B,C,D) -> f300(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = -1*x1 + x2 p(f2) = -1*x1 + x2 p(f300) = -1*x1 + x2 Following rules are strictly oriented: [B >= 1 + A] ==> f300(A,B,C,D) = -1*A + B > -1 + -1*A + B = f300(A,-1 + B,0,D) [0 >= 1 + E && B >= 1 + A] ==> f300(A,B,C,D) = -1*A + B > -1 + -1*A + B = f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A] ==> f300(A,B,C,D) = -1*A + B > -1 + -1*A + B = f300(1 + A,B,E,D) Following rules are weakly oriented: [A >= B] ==> f300(A,B,C,D) = -1*A + B >= -1*A + B = f1(A,B,C,E) True ==> f2(A,B,C,D) = -1*A + B >= -1*A + B = f300(A,B,C,D) * Step 5: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f300(A,B,C,D) -> f1(A,B,C,E) [A >= B] (1,1) 1. f300(A,B,C,D) -> f300(A,-1 + B,0,D) [B >= 1 + A] (A + B,1) 2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (A + B,1) 3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (A + B,1) 4. f2(A,B,C,D) -> f300(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))