The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
append(cons(x41_0, l42_0), l2) →⁺ cons(x41_0, append(l42_0, l2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [l42_0 / cons(x41_0, l42_0)].
The result substitution is [ ].

(2) BOUNDS(n^1, 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:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l1, l2, l1)
ifappend(l1, l2, nil) → l2
ifappend(l1, l2, cons(x, l)) → cons(x, append(l, l2))

S is empty.
Rewrite Strategy: FULL

(5) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
ifappend/0

(6) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l2, l1)
ifappend(l2, nil) → l2
ifappend(l2, cons(x, l)) → cons(x, append(l, l2))

S is empty.
Rewrite Strategy: FULL

(7) TypeInferenceProof (BOTH BOUNDS(ID, ID) transformation)

Infered types.

(8) Obligation:

TRS:
Rules:
is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l2, l1)
ifappend(l2, nil) → l2
ifappend(l2, cons(x, l)) → cons(x, append(l, l2))

Types:
is_empty :: nil:cons → true:false
nil :: nil:cons
true :: true:false
cons :: hd → nil:cons → nil:cons
false :: true:false
hd :: nil:cons → hd
tl :: nil:cons → nil:cons
append :: nil:cons → nil:cons → nil:cons
ifappend :: nil:cons → nil:cons → nil:cons
hole_true:false1_0 :: true:false
hole_nil:cons2_0 :: nil:cons
hole_hd3_0 :: hd
gen_nil:cons4_0 :: Nat → nil:cons

(9) OrderProof (LOWER BOUND(ID) transformation)

Heuristically decided to analyse the following defined symbols:
append, ifappend

They will be analysed ascendingly in the following order:
append = ifappend

(10) Obligation:

TRS:
Rules:
is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l2, l1)
ifappend(l2, nil) → l2
ifappend(l2, cons(x, l)) → cons(x, append(l, l2))

Types:
is_empty :: nil:cons → true:false
nil :: nil:cons
true :: true:false
cons :: hd → nil:cons → nil:cons
false :: true:false
hd :: nil:cons → hd
tl :: nil:cons → nil:cons
append :: nil:cons → nil:cons → nil:cons
ifappend :: nil:cons → nil:cons → nil:cons
hole_true:false1_0 :: true:false
hole_nil:cons2_0 :: nil:cons
hole_hd3_0 :: hd
gen_nil:cons4_0 :: Nat → nil:cons

Generator Equations:
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(hole_hd3_0, gen_nil:cons4_0(x))

The following defined symbols remain to be analysed:
ifappend, append

They will be analysed ascendingly in the following order:
append = ifappend

(11) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
ifappend(gen_nil:cons4_0(a), gen_nil:cons4_0(n6_0)) → gen_nil:cons4_0(+(n6_0, a)), rt ∈ Ω(1 + n6₀)

Induction Base:
ifappend(gen_nil:cons4_0(a), gen_nil:cons4_0(0)) →_R^Ω(1)
gen_nil:cons4_0(a)

Induction Step:
ifappend(gen_nil:cons4_0(a), gen_nil:cons4_0(+(n6_0, 1))) →_R^Ω(1)
cons(hole_hd3_0, append(gen_nil:cons4_0(n6_0), gen_nil:cons4_0(a))) →_R^Ω(1)
cons(hole_hd3_0, ifappend(gen_nil:cons4_0(a), gen_nil:cons4_0(n6_0))) →_IH
cons(hole_hd3_0, gen_nil:cons4_0(+(a, c7_0)))

We have rt ∈ Ω(n¹) and sz ∈ O(n). Thus, we have irc_R ∈ Ω(n).

(12) Complex Obligation (BEST)

(13) Obligation:

TRS:
Rules:
is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l2, l1)
ifappend(l2, nil) → l2
ifappend(l2, cons(x, l)) → cons(x, append(l, l2))

Types:
is_empty :: nil:cons → true:false
nil :: nil:cons
true :: true:false
cons :: hd → nil:cons → nil:cons
false :: true:false
hd :: nil:cons → hd
tl :: nil:cons → nil:cons
append :: nil:cons → nil:cons → nil:cons
ifappend :: nil:cons → nil:cons → nil:cons
hole_true:false1_0 :: true:false
hole_nil:cons2_0 :: nil:cons
hole_hd3_0 :: hd
gen_nil:cons4_0 :: Nat → nil:cons

Lemmas:
ifappend(gen_nil:cons4_0(a), gen_nil:cons4_0(n6_0)) → gen_nil:cons4_0(+(n6_0, a)), rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(hole_hd3_0, gen_nil:cons4_0(x))

The following defined symbols remain to be analysed:
append

They will be analysed ascendingly in the following order:
append = ifappend

(14) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

Could not prove a rewrite lemma for the defined symbol append.

(15) Obligation:

TRS:
Rules:
is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l2, l1)
ifappend(l2, nil) → l2
ifappend(l2, cons(x, l)) → cons(x, append(l, l2))

Types:
is_empty :: nil:cons → true:false
nil :: nil:cons
true :: true:false
cons :: hd → nil:cons → nil:cons
false :: true:false
hd :: nil:cons → hd
tl :: nil:cons → nil:cons
append :: nil:cons → nil:cons → nil:cons
ifappend :: nil:cons → nil:cons → nil:cons
hole_true:false1_0 :: true:false
hole_nil:cons2_0 :: nil:cons
hole_hd3_0 :: hd
gen_nil:cons4_0 :: Nat → nil:cons

Lemmas:
ifappend(gen_nil:cons4_0(a), gen_nil:cons4_0(n6_0)) → gen_nil:cons4_0(+(n6_0, a)), rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(hole_hd3_0, gen_nil:cons4_0(x))

No more defined symbols left to analyse.

(16) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
ifappend(gen_nil:cons4_0(a), gen_nil:cons4_0(n6_0)) → gen_nil:cons4_0(+(n6_0, a)), rt ∈ Ω(1 + n6₀)

(17) BOUNDS(n^1, INF)

(18) Obligation:

TRS:
Rules:
is_empty(nil) → true
is_empty(cons(x, l)) → false
hd(cons(x, l)) → x
tl(cons(x, l)) → l
append(l1, l2) → ifappend(l2, l1)
ifappend(l2, nil) → l2
ifappend(l2, cons(x, l)) → cons(x, append(l, l2))

Types:
is_empty :: nil:cons → true:false
nil :: nil:cons
true :: true:false
cons :: hd → nil:cons → nil:cons
false :: true:false
hd :: nil:cons → hd
tl :: nil:cons → nil:cons
append :: nil:cons → nil:cons → nil:cons
ifappend :: nil:cons → nil:cons → nil:cons
hole_true:false1_0 :: true:false
hole_nil:cons2_0 :: nil:cons
hole_hd3_0 :: hd
gen_nil:cons4_0 :: Nat → nil:cons

Lemmas:
ifappend(gen_nil:cons4_0(a), gen_nil:cons4_0(n6_0)) → gen_nil:cons4_0(+(n6_0, a)), rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_nil:cons4_0(0) ⇔ nil
gen_nil:cons4_0(+(x, 1)) ⇔ cons(hole_hd3_0, gen_nil:cons4_0(x))

No more defined symbols left to analyse.

(19) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
ifappend(gen_nil:cons4_0(a), gen_nil:cons4_0(n6_0)) → gen_nil:cons4_0(+(n6_0, a)), rt ∈ Ω(1 + n6₀)

(20) BOUNDS(n^1, INF)