WORST_CASE(?,O(n^1)) * Step 1: RestrictVarsProcessor WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A,B) -> f3(A,1) [0 >= A] (?,1) 1. f0(A,B) -> f2(A,0) True (1,1) 2. f2(A,B) -> f2(-1 + A,B) [A >= 1] (?,1) Signature: {(f0,2);(f2,2);(f3,2)} Flow Graph: [0->{},1->{0,2},2->{0,2}] + Applied Processor: RestrictVarsProcessor + Details: We removed the arguments [B] . * Step 2: LocalSizeboundsProc WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A) -> f3(A) [0 >= A] (?,1) 1. f0(A) -> f2(A) True (1,1) 2. f2(A) -> f2(-1 + A) [A >= 1] (?,1) Signature: {(f0,1);(f2,1);(f3,1)} Flow Graph: [0->{},1->{0,2},2->{0,2}] + Applied Processor: LocalSizeboundsProc + Details: LocalSizebounds generated; rvgraph (<0,0,A>, A, .= 0) (<1,0,A>, A, .= 0) (<2,0,A>, 1 + A, .+ 1) * Step 3: SizeboundsProc WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A) -> f3(A) [0 >= A] (?,1) 1. f0(A) -> f2(A) True (1,1) 2. f2(A) -> f2(-1 + A) [A >= 1] (?,1) Signature: {(f0,1);(f2,1);(f3,1)} Flow Graph: [0->{},1->{0,2},2->{0,2}] Sizebounds: (<0,0,A>, ?) (<1,0,A>, ?) (<2,0,A>, ?) + Applied Processor: SizeboundsProc + Details: Sizebounds computed: (<0,0,A>, ?) (<1,0,A>, A) (<2,0,A>, ?) * Step 4: LeafRules WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A) -> f3(A) [0 >= A] (?,1) 1. f0(A) -> f2(A) True (1,1) 2. f2(A) -> f2(-1 + A) [A >= 1] (?,1) Signature: {(f0,1);(f2,1);(f3,1)} Flow Graph: [0->{},1->{0,2},2->{0,2}] Sizebounds: (<0,0,A>, ?) (<1,0,A>, A) (<2,0,A>, ?) + Applied Processor: LeafRules + Details: The following transitions are estimated by its predecessors and are removed [0] * Step 5: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 1. f0(A) -> f2(A) True (1,1) 2. f2(A) -> f2(-1 + A) [A >= 1] (?,1) Signature: {(f0,1);(f2,1);(f3,1)} Flow Graph: [1->{2},2->{2}] Sizebounds: (<1,0,A>, A) (<2,0,A>, ?) + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = x1 p(f2) = x1 The following rules are strictly oriented: [A >= 1] ==> f2(A) = A > -1 + A = f2(-1 + A) The following rules are weakly oriented: True ==> f0(A) = A >= A = f2(A) * Step 6: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 1. f0(A) -> f2(A) True (1,1) 2. f2(A) -> f2(-1 + A) [A >= 1] (A,1) Signature: {(f0,1);(f2,1);(f3,1)} Flow Graph: [1->{2},2->{2}] Sizebounds: (<1,0,A>, A) (<2,0,A>, ?) + Applied Processor: KnowledgePropagation + Details: The problem is already solved. WORST_CASE(?,O(n^1))