*** 1 Progress [(O(1),O(n^1))] *** Considered Problem: Strict DP Rules: Strict TRS Rules: activate(X) -> X and(tt(),X) -> activate(X) plus(N,0()) -> N plus(N,s(M)) -> s(plus(N,M)) Weak DP Rules: Weak TRS Rules: Signature: {activate/1,and/2,plus/2} / {0/0,s/1,tt/0} Obligation: Full basic terms: {activate,and,plus}/{0,s,tt} Applied Processor: ToInnermost Proof: switch to innermost, as the system is overlay and right linear and does not contain weak rules *** 1.1 Progress [(O(1),O(n^1))] *** Considered Problem: Strict DP Rules: Strict TRS Rules: activate(X) -> X and(tt(),X) -> activate(X) plus(N,0()) -> N plus(N,s(M)) -> s(plus(N,M)) Weak DP Rules: Weak TRS Rules: Signature: {activate/1,and/2,plus/2} / {0/0,s/1,tt/0} Obligation: Innermost basic terms: {activate,and,plus}/{0,s,tt} Applied Processor: Bounds {initialAutomaton = perSymbol, enrichment = match} Proof: The problem is match-bounded by 1. The enriched problem is compatible with follwoing automaton. 0_0() -> 1 0_0() -> 2 0_0() -> 3 0_0() -> 4 0_0() -> 7 activate_0(1) -> 2 activate_0(5) -> 2 activate_0(6) -> 2 activate_1(1) -> 3 activate_1(5) -> 3 activate_1(6) -> 3 and_0(1,1) -> 3 and_0(1,5) -> 3 and_0(1,6) -> 3 and_0(5,1) -> 3 and_0(5,5) -> 3 and_0(5,6) -> 3 and_0(6,1) -> 3 and_0(6,5) -> 3 and_0(6,6) -> 3 plus_0(1,1) -> 4 plus_0(1,5) -> 4 plus_0(1,6) -> 4 plus_0(5,1) -> 4 plus_0(5,5) -> 4 plus_0(5,6) -> 4 plus_0(6,1) -> 4 plus_0(6,5) -> 4 plus_0(6,6) -> 4 plus_1(1,1) -> 7 plus_1(1,5) -> 7 plus_1(1,6) -> 7 plus_1(5,1) -> 7 plus_1(5,5) -> 7 plus_1(5,6) -> 7 plus_1(6,1) -> 7 plus_1(6,5) -> 7 plus_1(6,6) -> 7 s_0(1) -> 2 s_0(1) -> 3 s_0(1) -> 4 s_0(1) -> 5 s_0(1) -> 7 s_0(5) -> 2 s_0(5) -> 3 s_0(5) -> 4 s_0(5) -> 5 s_0(5) -> 7 s_0(6) -> 2 s_0(6) -> 3 s_0(6) -> 4 s_0(6) -> 5 s_0(6) -> 7 s_1(7) -> 4 s_1(7) -> 7 tt_0() -> 2 tt_0() -> 3 tt_0() -> 4 tt_0() -> 6 tt_0() -> 7 1 -> 2 1 -> 3 1 -> 4 1 -> 7 5 -> 2 5 -> 3 5 -> 4 5 -> 7 6 -> 2 6 -> 3 6 -> 4 6 -> 7 *** 1.1.1 Progress [(O(1),O(1))] *** Considered Problem: Strict DP Rules: Strict TRS Rules: Weak DP Rules: Weak TRS Rules: activate(X) -> X and(tt(),X) -> activate(X) plus(N,0()) -> N plus(N,s(M)) -> s(plus(N,M)) Signature: {activate/1,and/2,plus/2} / {0/0,s/1,tt/0} Obligation: Innermost basic terms: {activate,and,plus}/{0,s,tt} Applied Processor: EmptyProcessor Proof: The problem is already closed. The intended complexity is O(1).