(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
f(g(x)) → f(a(g(g(f(x))), g(f(x))))
Rewrite Strategy: INNERMOST
(1) DecreasingLoopProof (EQUIVALENT transformation)
The following loop(s) give(s) rise to the lower bound Ω(2n):
The rewrite sequence
f(g(x)) →+ f(a(g(g(f(x))), g(f(x))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,0,0].
The pumping substitution is [x / g(x)].
The result substitution is [ ].
The rewrite sequence
f(g(x)) →+ f(a(g(g(f(x))), g(f(x))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0].
The pumping substitution is [x / g(x)].
The result substitution is [ ].
(2) BOUNDS(2^n, INF)
(3) RenamingProof (EQUIVALENT transformation)
Renamed function symbols to avoid clashes with predefined symbol.
(4) Obligation:
Runtime Complexity Relative TRS:
The TRS R consists of the following rules:
f(g(x)) → f(a(g(g(f(x))), g(f(x))))
S is empty.
Rewrite Strategy: INNERMOST
(5) TypeInferenceProof (BOTH BOUNDS(ID, ID) transformation)
Infered types.
(6) Obligation:
Innermost TRS:
Rules:
f(g(x)) → f(a(g(g(f(x))), g(f(x))))
Types:
f :: g:a → g:a
g :: g:a → g:a
a :: g:a → g:a → g:a
hole_g:a1_0 :: g:a
gen_g:a2_0 :: Nat → g:a
(7) OrderProof (LOWER BOUND(ID) transformation)
Heuristically decided to analyse the following defined symbols:
f
(8) Obligation:
Innermost TRS:
Rules:
f(
g(
x)) →
f(
a(
g(
g(
f(
x))),
g(
f(
x))))
Types:
f :: g:a → g:a
g :: g:a → g:a
a :: g:a → g:a → g:a
hole_g:a1_0 :: g:a
gen_g:a2_0 :: Nat → g:a
Generator Equations:
gen_g:a2_0(0) ⇔ hole_g:a1_0
gen_g:a2_0(+(x, 1)) ⇔ g(gen_g:a2_0(x))
The following defined symbols remain to be analysed:
f
(9) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)
Could not prove a rewrite lemma for the defined symbol f.
(10) Obligation:
Innermost TRS:
Rules:
f(
g(
x)) →
f(
a(
g(
g(
f(
x))),
g(
f(
x))))
Types:
f :: g:a → g:a
g :: g:a → g:a
a :: g:a → g:a → g:a
hole_g:a1_0 :: g:a
gen_g:a2_0 :: Nat → g:a
Generator Equations:
gen_g:a2_0(0) ⇔ hole_g:a1_0
gen_g:a2_0(+(x, 1)) ⇔ g(gen_g:a2_0(x))
No more defined symbols left to analyse.