YES(?,O(n^1)) * Step 1: ArgumentFilter WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,A1) -> f2(A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,A1) True (1,1) 1. f2(A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,A1) -> f10(A,B,C,1,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,A1) [B >= 1 + A] (?,1) 2. f2(A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,A1) -> f2(A,1 + B,B1*C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,A1) [A >= B] (?,1) 3. f10(A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,A1) -> f1(A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,A1) [1 + -1*D >= 0 && -1 + D >= 0 && -1 + -1*A + B >= 0 && 0 >= B] (?,1) Signature: {(f1,27);(f10,27);(f2,27);(start,27)} Flow Graph: [0->{1,2},1->{3},2->{1,2},3->{}] + Applied Processor: ArgumentFilter [2,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26] + Details: We remove following argument positions: [2,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26]. * Step 2: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,D) -> f2(A,B,D) True (1,1) 1. f2(A,B,D) -> f10(A,B,1) [B >= 1 + A] (?,1) 2. f2(A,B,D) -> f2(A,1 + B,D) [A >= B] (?,1) 3. f10(A,B,D) -> f1(A,B,D) [1 + -1*D >= 0 && -1 + D >= 0 && -1 + -1*A + B >= 0 && 0 >= B] (?,1) Signature: {(f1,27);(f10,27);(f2,27);(start,27)} Flow Graph: [0->{1,2},1->{3},2->{1,2},3->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 3: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,D) -> f2(A,B,D) True (1,1) 1. f2(A,B,D) -> f10(A,B,1) [B >= 1 + A] (1,1) 2. f2(A,B,D) -> f2(A,1 + B,D) [A >= B] (?,1) 3. f10(A,B,D) -> f1(A,B,D) [1 + -1*D >= 0 && -1 + D >= 0 && -1 + -1*A + B >= 0 && 0 >= B] (1,1) Signature: {(f1,27);(f10,27);(f2,27);(start,27)} Flow Graph: [0->{1,2},1->{3},2->{1,2},3->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = x1 + -1*x2 + x27 p(f10) = x1 + -1*x2 + x27 p(f2) = x1 + -1*x2 + x27 p(start) = x1 + -1*x2 + x27 Following rules are strictly oriented: [A >= B] ==> f2(A,B,D) = 1 + A + -1*B > A + -1*B = f2(A,1 + B,D) Following rules are weakly oriented: True ==> start(A,B,D) = 1 + A + -1*B >= 1 + A + -1*B = f2(A,B,D) [B >= 1 + A] ==> f2(A,B,D) = 1 + A + -1*B >= 1 + A + -1*B = f10(A,B,1) [1 + -1*D >= 0 && -1 + D >= 0 && -1 + -1*A + B >= 0 && 0 >= B] ==> f10(A,B,D) = 1 + A + -1*B >= 1 + A + -1*B = f1(A,B,D) * Step 4: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. start(A,B,D) -> f2(A,B,D) True (1,1) 1. f2(A,B,D) -> f10(A,B,1) [B >= 1 + A] (1,1) 2. f2(A,B,D) -> f2(A,1 + B,D) [A >= B] (1 + A + B,1) 3. f10(A,B,D) -> f1(A,B,D) [1 + -1*D >= 0 && -1 + D >= 0 && -1 + -1*A + B >= 0 && 0 >= B] (1,1) Signature: {(f1,27);(f10,27);(f2,27);(start,27)} Flow Graph: [0->{1,2},1->{3},2->{1,2},3->{}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))