YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [E + -1*F >= 0 (?,1) && -1*E + F >= 0 && C + -1*D >= 0 && -1*C + D >= 0 && A + -1*B >= 0 && -1*A + B >= 0 && 0 >= A && B = A && C = D && E = F] 1. start(A,B,C,D,E,F) -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [E + -1*F >= 0 (?,1) && -1*E + F >= 0 && C + -1*D >= 0 && -1*C + D >= 0 && A + -1*B >= 0 && -1*A + B >= 0 && A >= 1 && B = A && C = D && E = F] 2. lbl71(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [-1 + E + -1*F >= 0 (?,1) && -1 + -1*C + D >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && D >= 1 + C && B = 0 && C + E = D + F && A + C = D] 3. lbl71(A,B,C,D,E,F) -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [-1 + E + -1*F >= 0 (?,1) && -1 + -1*C + D >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && A + C >= 1 + D && D >= 1 + C && A + C >= D && C + E = D + F && B + D = A + C] 4. start0(A,B,C,D,E,F) -> start(A,A,D,D,F,F) True (1,1) Signature: {(lbl71,6);(start,6);(start0,6);(stop,6)} Flow Graph: [0->{},1->{2,3},2->{},3->{2,3},4->{0,1}] + 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. start(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [E + -1*F >= 0 (1,1) && -1*E + F >= 0 && C + -1*D >= 0 && -1*C + D >= 0 && A + -1*B >= 0 && -1*A + B >= 0 && 0 >= A && B = A && C = D && E = F] 1. start(A,B,C,D,E,F) -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [E + -1*F >= 0 (1,1) && -1*E + F >= 0 && C + -1*D >= 0 && -1*C + D >= 0 && A + -1*B >= 0 && -1*A + B >= 0 && A >= 1 && B = A && C = D && E = F] 2. lbl71(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [-1 + E + -1*F >= 0 (1,1) && -1 + -1*C + D >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && D >= 1 + C && B = 0 && C + E = D + F && A + C = D] 3. lbl71(A,B,C,D,E,F) -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [-1 + E + -1*F >= 0 (?,1) && -1 + -1*C + D >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && A + C >= 1 + D && D >= 1 + C && A + C >= D && C + E = D + F && B + D = A + C] 4. start0(A,B,C,D,E,F) -> start(A,A,D,D,F,F) True (1,1) Signature: {(lbl71,6);(start,6);(start0,6);(stop,6)} Flow Graph: [0->{},1->{2,3},2->{},3->{2,3},4->{0,1}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(lbl71) = 1 + x1 + x3 + -1*x4 p(start) = x1 + x3 + -1*x4 p(start0) = x1 p(stop) = x1 + x3 + -1*x4 Following rules are strictly oriented: [-1 + E + -1*F >= 0 ==> && -1 + -1*C + D >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && D >= 1 + C && B = 0 && C + E = D + F && A + C = D] lbl71(A,B,C,D,E,F) = 1 + A + C + -1*D > A + C + -1*D = stop(A,B,C,D,E,F) [-1 + E + -1*F >= 0 ==> && -1 + -1*C + D >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && A + C >= 1 + D && D >= 1 + C && A + C >= D && C + E = D + F && B + D = A + C] lbl71(A,B,C,D,E,F) = 1 + A + C + -1*D > A + C + -1*D = lbl71(A,-1 + B,-1 + C,D,1 + E,F) Following rules are weakly oriented: [E + -1*F >= 0 ==> && -1*E + F >= 0 && C + -1*D >= 0 && -1*C + D >= 0 && A + -1*B >= 0 && -1*A + B >= 0 && 0 >= A && B = A && C = D && E = F] start(A,B,C,D,E,F) = A + C + -1*D >= A + C + -1*D = stop(A,B,C,D,E,F) [E + -1*F >= 0 ==> && -1*E + F >= 0 && C + -1*D >= 0 && -1*C + D >= 0 && A + -1*B >= 0 && -1*A + B >= 0 && A >= 1 && B = A && C = D && E = F] start(A,B,C,D,E,F) = A + C + -1*D >= A + C + -1*D = lbl71(A,-1 + B,-1 + C,D,1 + E,F) True ==> start0(A,B,C,D,E,F) = A >= A = start(A,A,D,D,F,F) * Step 3: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [E + -1*F >= 0 (1,1) && -1*E + F >= 0 && C + -1*D >= 0 && -1*C + D >= 0 && A + -1*B >= 0 && -1*A + B >= 0 && 0 >= A && B = A && C = D && E = F] 1. start(A,B,C,D,E,F) -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [E + -1*F >= 0 (1,1) && -1*E + F >= 0 && C + -1*D >= 0 && -1*C + D >= 0 && A + -1*B >= 0 && -1*A + B >= 0 && A >= 1 && B = A && C = D && E = F] 2. lbl71(A,B,C,D,E,F) -> stop(A,B,C,D,E,F) [-1 + E + -1*F >= 0 (1,1) && -1 + -1*C + D >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && D >= 1 + C && B = 0 && C + E = D + F && A + C = D] 3. lbl71(A,B,C,D,E,F) -> lbl71(A,-1 + B,-1 + C,D,1 + E,F) [-1 + E + -1*F >= 0 (A,1) && -1 + -1*C + D >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && A + C >= 1 + D && D >= 1 + C && A + C >= D && C + E = D + F && B + D = A + C] 4. start0(A,B,C,D,E,F) -> start(A,A,D,D,F,F) True (1,1) Signature: {(lbl71,6);(start,6);(start0,6);(stop,6)} Flow Graph: [0->{},1->{2,3},2->{},3->{2,3},4->{0,1}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))