*** 1 Progress [(O(1),O(n^1))] ***
Considered Problem:
Strict DP Rules:
Strict TRS Rules:
0() -> n__0()
activate(X) -> X
activate(n__0()) -> 0()
activate(n__f(X)) -> f(activate(X))
activate(n__s(X)) -> s(activate(X))
f(X) -> n__f(X)
f(0()) -> cons(0(),n__f(n__s(n__0())))
f(s(0())) -> f(p(s(0())))
p(s(X)) -> X
s(X) -> n__s(X)
Weak DP Rules:
Weak TRS Rules:
Signature:
{0/0,activate/1,f/1,p/1,s/1} / {cons/2,n__0/0,n__f/1,n__s/1}
Obligation:
Innermost
basic terms: {0,activate,f,p,s}/{cons,n__0,n__f,n__s}
Applied Processor:
InnermostRuleRemoval
Proof:
Arguments of following rules are not normal-forms.
f(0()) -> cons(0(),n__f(n__s(n__0())))
f(s(0())) -> f(p(s(0())))
p(s(X)) -> X
All above mentioned rules can be savely removed.
*** 1.1 Progress [(O(1),O(n^1))] ***
Considered Problem:
Strict DP Rules:
Strict TRS Rules:
0() -> n__0()
activate(X) -> X
activate(n__0()) -> 0()
activate(n__f(X)) -> f(activate(X))
activate(n__s(X)) -> s(activate(X))
f(X) -> n__f(X)
s(X) -> n__s(X)
Weak DP Rules:
Weak TRS Rules:
Signature:
{0/0,activate/1,f/1,p/1,s/1} / {cons/2,n__0/0,n__f/1,n__s/1}
Obligation:
Innermost
basic terms: {0,activate,f,p,s}/{cons,n__0,n__f,n__s}
Applied Processor:
Bounds {initialAutomaton = minimal, enrichment = match}
Proof:
The problem is match-bounded by 2.
The enriched problem is compatible with follwoing automaton.
0_0() -> 1
0_1() -> 1
0_1() -> 3
activate_0(2) -> 1
activate_1(2) -> 3
cons_0(2,2) -> 1
cons_0(2,2) -> 2
cons_0(2,2) -> 3
f_0(2) -> 1
f_1(3) -> 1
f_1(3) -> 3
n__0_0() -> 1
n__0_0() -> 2
n__0_0() -> 3
n__0_1() -> 1
n__0_2() -> 1
n__0_2() -> 3
n__f_0(2) -> 1
n__f_0(2) -> 2
n__f_0(2) -> 3
n__f_1(2) -> 1
n__f_2(3) -> 1
n__f_2(3) -> 3
n__s_0(2) -> 1
n__s_0(2) -> 2
n__s_0(2) -> 3
n__s_1(2) -> 1
n__s_2(3) -> 1
n__s_2(3) -> 3
p_0(2) -> 1
s_0(2) -> 1
s_1(3) -> 1
s_1(3) -> 3
2 -> 1
2 -> 3
*** 1.1.1 Progress [(O(1),O(1))] ***
Considered Problem:
Strict DP Rules:
Strict TRS Rules:
Weak DP Rules:
Weak TRS Rules:
0() -> n__0()
activate(X) -> X
activate(n__0()) -> 0()
activate(n__f(X)) -> f(activate(X))
activate(n__s(X)) -> s(activate(X))
f(X) -> n__f(X)
s(X) -> n__s(X)
Signature:
{0/0,activate/1,f/1,p/1,s/1} / {cons/2,n__0/0,n__f/1,n__s/1}
Obligation:
Innermost
basic terms: {0,activate,f,p,s}/{cons,n__0,n__f,n__s}
Applied Processor:
EmptyProcessor
Proof:
The problem is already closed. The intended complexity is O(1).