MAYBE * Step 1: UnsatPaths MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (?,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (?,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (?,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) Signature: {(f2,3);(f3,3);(f300,3)} 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: UnsatPaths + Details: We remove following edges from the transition graph: [(1,0)] * Step 2: TrivialSCCs MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (?,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (?,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (?,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) Signature: {(f2,3);(f3,3);(f300,3)} Flow Graph: [0->{},1->{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 3: PolyRank MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (1,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (?,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (?,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) Signature: {(f2,3);(f3,3);(f300,3)} Flow Graph: [0->{},1->{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(f2) = 1 + x1 p(f3) = 1 + x1 p(f300) = 1 + x1 Following rules are strictly oriented: [D >= 1 && A >= 0] ==> f2(A,B,C) = 1 + A > A = f2(-1 + A,D,C) Following rules are weakly oriented: [0 >= 1 + A] ==> f2(A,B,C) = 1 + A >= 1 + A = f300(A,B,D) [A >= 0] ==> f2(A,B,C) = 1 + A >= 1 + A = f2(A,0,C) [0 >= 1 + D && A >= 0] ==> f2(A,B,C) = 1 + A >= A = f2(-1 + A,D,C) True ==> f3(A,B,C) = 1 + A >= 1 + A = f2(A,B,C) * Step 4: PolyRank MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (1,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (?,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (1 + A,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) Signature: {(f2,3);(f3,3);(f300,3)} Flow Graph: [0->{},1->{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(f2) = 1 + x1 p(f3) = 1 + x1 p(f300) = 1 + x1 Following rules are strictly oriented: [0 >= 1 + D && A >= 0] ==> f2(A,B,C) = 1 + A > A = f2(-1 + A,D,C) [D >= 1 && A >= 0] ==> f2(A,B,C) = 1 + A > A = f2(-1 + A,D,C) Following rules are weakly oriented: [0 >= 1 + A] ==> f2(A,B,C) = 1 + A >= 1 + A = f300(A,B,D) [A >= 0] ==> f2(A,B,C) = 1 + A >= 1 + A = f2(A,0,C) True ==> f3(A,B,C) = 1 + A >= 1 + A = f2(A,B,C) * Step 5: Failure MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (1,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (1 + A,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (1 + A,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) Signature: {(f2,3);(f3,3);(f300,3)} Flow Graph: [0->{},1->{1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}] + Applied Processor: Failing "Open problems left." + Details: Open problems left. MAYBE