MAYBE * Step 1: UnsatPaths MAYBE + Considered Problem: Rules: 0. f1(A,B,C,D) -> f300(A,B,C,E) [A >= B && A >= 1 + B] (?,1) 1. f1(A,B,C,D) -> f1(A,A,0,D) [B >= E && A = B] (?,1) 2. f1(A,B,C,D) -> f1(1 + A,A,E,D) [0 >= 1 + E && B >= F && A = B] (?,1) 3. f1(A,B,C,D) -> f1(1 + A,A,E,D) [E >= 1 && B >= F && A = B] (?,1) 4. f1(A,B,C,D) -> f1(A,B,0,D) [B >= 1 + A] (?,1) 5. f1(A,B,C,D) -> f1(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1) 6. f1(A,B,C,D) -> f1(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (?,1) 7. f2(A,B,C,D) -> f1(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{0,1,2,3,4,5,6},2->{0,1,2,3,4,5,6},3->{0,1,2,3,4,5,6},4->{0,1,2,3,4,5,6},5->{0,1,2,3,4,5,6} ,6->{0,1,2,3,4,5,6},7->{0,1,2,3,4,5,6}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(1,0) ,(1,4) ,(1,5) ,(1,6) ,(2,1) ,(2,2) ,(2,3) ,(2,4) ,(2,5) ,(2,6) ,(3,1) ,(3,2) ,(3,3) ,(3,4) ,(3,5) ,(3,6) ,(4,0) ,(4,1) ,(4,2) ,(4,3) ,(5,0) ,(6,0)] * Step 2: TrivialSCCs MAYBE + Considered Problem: Rules: 0. f1(A,B,C,D) -> f300(A,B,C,E) [A >= B && A >= 1 + B] (?,1) 1. f1(A,B,C,D) -> f1(A,A,0,D) [B >= E && A = B] (?,1) 2. f1(A,B,C,D) -> f1(1 + A,A,E,D) [0 >= 1 + E && B >= F && A = B] (?,1) 3. f1(A,B,C,D) -> f1(1 + A,A,E,D) [E >= 1 && B >= F && A = B] (?,1) 4. f1(A,B,C,D) -> f1(A,B,0,D) [B >= 1 + A] (?,1) 5. f1(A,B,C,D) -> f1(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1) 6. f1(A,B,C,D) -> f1(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (?,1) 7. f2(A,B,C,D) -> f1(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{1,2,3},2->{0},3->{0},4->{4,5,6},5->{1,2,3,4,5,6},6->{1,2,3,4,5,6},7->{0,1,2,3,4,5,6}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 3: PolyRank MAYBE + Considered Problem: Rules: 0. f1(A,B,C,D) -> f300(A,B,C,E) [A >= B && A >= 1 + B] (1,1) 1. f1(A,B,C,D) -> f1(A,A,0,D) [B >= E && A = B] (?,1) 2. f1(A,B,C,D) -> f1(1 + A,A,E,D) [0 >= 1 + E && B >= F && A = B] (1,1) 3. f1(A,B,C,D) -> f1(1 + A,A,E,D) [E >= 1 && B >= F && A = B] (1,1) 4. f1(A,B,C,D) -> f1(A,B,0,D) [B >= 1 + A] (?,1) 5. f1(A,B,C,D) -> f1(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1) 6. f1(A,B,C,D) -> f1(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (?,1) 7. f2(A,B,C,D) -> f1(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{1,2,3},2->{0},3->{0},4->{4,5,6},5->{1,2,3,4,5,6},6->{1,2,3,4,5,6},7->{0,1,2,3,4,5,6}] + 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] ==> f1(A,B,C,D) = -1*A + B > -1 + -1*A + B = f1(1 + A,B,E,D) Following rules are weakly oriented: [A >= B && A >= 1 + B] ==> f1(A,B,C,D) = -1*A + B >= -1*A + B = f300(A,B,C,E) [B >= E && A = B] ==> f1(A,B,C,D) = -1*A + B >= 0 = f1(A,A,0,D) [0 >= 1 + E && B >= F && A = B] ==> f1(A,B,C,D) = -1*A + B >= -1 = f1(1 + A,A,E,D) [E >= 1 && B >= F && A = B] ==> f1(A,B,C,D) = -1*A + B >= -1 = f1(1 + A,A,E,D) [B >= 1 + A] ==> f1(A,B,C,D) = -1*A + B >= -1*A + B = f1(A,B,0,D) [0 >= 1 + E && B >= 1 + A] ==> f1(A,B,C,D) = -1*A + B >= -1 + -1*A + B = f1(1 + A,B,E,D) True ==> f2(A,B,C,D) = -1*A + B >= -1*A + B = f1(A,B,C,D) * Step 4: PolyRank MAYBE + Considered Problem: Rules: 0. f1(A,B,C,D) -> f300(A,B,C,E) [A >= B && A >= 1 + B] (1,1) 1. f1(A,B,C,D) -> f1(A,A,0,D) [B >= E && A = B] (?,1) 2. f1(A,B,C,D) -> f1(1 + A,A,E,D) [0 >= 1 + E && B >= F && A = B] (1,1) 3. f1(A,B,C,D) -> f1(1 + A,A,E,D) [E >= 1 && B >= F && A = B] (1,1) 4. f1(A,B,C,D) -> f1(A,B,0,D) [B >= 1 + A] (?,1) 5. f1(A,B,C,D) -> f1(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1) 6. f1(A,B,C,D) -> f1(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (A + B,1) 7. f2(A,B,C,D) -> f1(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{1,2,3},2->{0},3->{0},4->{4,5,6},5->{1,2,3,4,5,6},6->{1,2,3,4,5,6},7->{0,1,2,3,4,5,6}] + 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] ==> f1(A,B,C,D) = -1*A + B > -1 + -1*A + B = f1(1 + A,B,E,D) [E >= 1 && B >= 1 + A] ==> f1(A,B,C,D) = -1*A + B > -1 + -1*A + B = f1(1 + A,B,E,D) Following rules are weakly oriented: [A >= B && A >= 1 + B] ==> f1(A,B,C,D) = -1*A + B >= -1*A + B = f300(A,B,C,E) [B >= E && A = B] ==> f1(A,B,C,D) = -1*A + B >= 0 = f1(A,A,0,D) [0 >= 1 + E && B >= F && A = B] ==> f1(A,B,C,D) = -1*A + B >= -1 = f1(1 + A,A,E,D) [E >= 1 && B >= F && A = B] ==> f1(A,B,C,D) = -1*A + B >= -1 = f1(1 + A,A,E,D) [B >= 1 + A] ==> f1(A,B,C,D) = -1*A + B >= -1*A + B = f1(A,B,0,D) True ==> f2(A,B,C,D) = -1*A + B >= -1*A + B = f1(A,B,C,D) * Step 5: Failure MAYBE + Considered Problem: Rules: 0. f1(A,B,C,D) -> f300(A,B,C,E) [A >= B && A >= 1 + B] (1,1) 1. f1(A,B,C,D) -> f1(A,A,0,D) [B >= E && A = B] (?,1) 2. f1(A,B,C,D) -> f1(1 + A,A,E,D) [0 >= 1 + E && B >= F && A = B] (1,1) 3. f1(A,B,C,D) -> f1(1 + A,A,E,D) [E >= 1 && B >= F && A = B] (1,1) 4. f1(A,B,C,D) -> f1(A,B,0,D) [B >= 1 + A] (?,1) 5. f1(A,B,C,D) -> f1(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (A + B,1) 6. f1(A,B,C,D) -> f1(1 + A,B,E,D) [E >= 1 && B >= 1 + A] (A + B,1) 7. f2(A,B,C,D) -> f1(A,B,C,D) True (1,1) Signature: {(f1,4);(f2,4);(f300,4)} Flow Graph: [0->{},1->{1,2,3},2->{0},3->{0},4->{4,5,6},5->{1,2,3,4,5,6},6->{1,2,3,4,5,6},7->{0,1,2,3,4,5,6}] + Applied Processor: Failing "Open problems left." + Details: Open problems left. MAYBE