### (0) Obligation:

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

numbersd(0)
d(x) → if(le(x, nr), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0)))))), 0)
ack(0, x) → s(x)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Rewrite Strategy: FULL

### (1) DecreasingLoopProof (EQUIVALENT transformation)

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

### (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:

numbersd(0')
d(x) → if(le(x, nr), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0', y) → true
le(s(x), 0') → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0')))))), 0')
ack(0', x) → s(x)
ack(s(x), 0') → ack(x, s(0'))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

S is empty.
Rewrite Strategy: FULL

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

Sliced the following arguments:
cons/0

### (6) Obligation:

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

numbersd(0')
d(x) → if(le(x, nr), x)
if(true, x) → cons(d(s(x)))
if(false, x) → nil
le(0', y) → true
le(s(x), 0') → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0')))))), 0')
ack(0', x) → s(x)
ack(s(x), 0') → ack(x, s(0'))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

S is empty.
Rewrite Strategy: FULL

Infered types.

### (8) Obligation:

TRS:
Rules:
numbersd(0')
d(x) → if(le(x, nr), x)
if(true, x) → cons(d(s(x)))
if(false, x) → nil
le(0', y) → true
le(s(x), 0') → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0')))))), 0')
ack(0', x) → s(x)
ack(s(x), 0') → ack(x, s(0'))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Types:
numbers :: cons:nil
d :: 0':s → cons:nil
0' :: 0':s
if :: true:false → 0':s → cons:nil
le :: 0':s → 0':s → true:false
nr :: 0':s
true :: true:false
cons :: cons:nil → cons:nil
s :: 0':s → 0':s
false :: true:false
nil :: cons:nil
ack :: 0':s → 0':s → 0':s
hole_cons:nil1_0 :: cons:nil
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_cons:nil4_0 :: Nat → cons:nil
gen_0':s5_0 :: Nat → 0':s

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

Heuristically decided to analyse the following defined symbols:
d, le, ack

They will be analysed ascendingly in the following order:
le < d

### (10) Obligation:

TRS:
Rules:
numbersd(0')
d(x) → if(le(x, nr), x)
if(true, x) → cons(d(s(x)))
if(false, x) → nil
le(0', y) → true
le(s(x), 0') → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0')))))), 0')
ack(0', x) → s(x)
ack(s(x), 0') → ack(x, s(0'))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Types:
numbers :: cons:nil
d :: 0':s → cons:nil
0' :: 0':s
if :: true:false → 0':s → cons:nil
le :: 0':s → 0':s → true:false
nr :: 0':s
true :: true:false
cons :: cons:nil → cons:nil
s :: 0':s → 0':s
false :: true:false
nil :: cons:nil
ack :: 0':s → 0':s → 0':s
hole_cons:nil1_0 :: cons:nil
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_cons:nil4_0 :: Nat → cons:nil
gen_0':s5_0 :: Nat → 0':s

Generator Equations:
gen_cons:nil4_0(0) ⇔ nil
gen_cons:nil4_0(+(x, 1)) ⇔ cons(gen_cons:nil4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))

The following defined symbols remain to be analysed:
le, d, ack

They will be analysed ascendingly in the following order:
le < d

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

Proved the following rewrite lemma:
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) → true, rt ∈ Ω(1 + n70)

Induction Base:
le(gen_0':s5_0(0), gen_0':s5_0(0)) →RΩ(1)
true

Induction Step:
le(gen_0':s5_0(+(n7_0, 1)), gen_0':s5_0(+(n7_0, 1))) →RΩ(1)
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) →IH
true

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

### (13) Obligation:

TRS:
Rules:
numbersd(0')
d(x) → if(le(x, nr), x)
if(true, x) → cons(d(s(x)))
if(false, x) → nil
le(0', y) → true
le(s(x), 0') → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0')))))), 0')
ack(0', x) → s(x)
ack(s(x), 0') → ack(x, s(0'))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Types:
numbers :: cons:nil
d :: 0':s → cons:nil
0' :: 0':s
if :: true:false → 0':s → cons:nil
le :: 0':s → 0':s → true:false
nr :: 0':s
true :: true:false
cons :: cons:nil → cons:nil
s :: 0':s → 0':s
false :: true:false
nil :: cons:nil
ack :: 0':s → 0':s → 0':s
hole_cons:nil1_0 :: cons:nil
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_cons:nil4_0 :: Nat → cons:nil
gen_0':s5_0 :: Nat → 0':s

Lemmas:
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) → true, rt ∈ Ω(1 + n70)

Generator Equations:
gen_cons:nil4_0(0) ⇔ nil
gen_cons:nil4_0(+(x, 1)) ⇔ cons(gen_cons:nil4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))

The following defined symbols remain to be analysed:
d, ack

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

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

### (15) Obligation:

TRS:
Rules:
numbersd(0')
d(x) → if(le(x, nr), x)
if(true, x) → cons(d(s(x)))
if(false, x) → nil
le(0', y) → true
le(s(x), 0') → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0')))))), 0')
ack(0', x) → s(x)
ack(s(x), 0') → ack(x, s(0'))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Types:
numbers :: cons:nil
d :: 0':s → cons:nil
0' :: 0':s
if :: true:false → 0':s → cons:nil
le :: 0':s → 0':s → true:false
nr :: 0':s
true :: true:false
cons :: cons:nil → cons:nil
s :: 0':s → 0':s
false :: true:false
nil :: cons:nil
ack :: 0':s → 0':s → 0':s
hole_cons:nil1_0 :: cons:nil
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_cons:nil4_0 :: Nat → cons:nil
gen_0':s5_0 :: Nat → 0':s

Lemmas:
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) → true, rt ∈ Ω(1 + n70)

Generator Equations:
gen_cons:nil4_0(0) ⇔ nil
gen_cons:nil4_0(+(x, 1)) ⇔ cons(gen_cons:nil4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))

The following defined symbols remain to be analysed:
ack

### (16) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
ack(gen_0':s5_0(1), gen_0':s5_0(+(1, n934_0))) → *6_0, rt ∈ Ω(n9340)

Induction Base:
ack(gen_0':s5_0(1), gen_0':s5_0(+(1, 0)))

Induction Step:
ack(gen_0':s5_0(1), gen_0':s5_0(+(1, +(n934_0, 1)))) →RΩ(1)
ack(gen_0':s5_0(0), ack(s(gen_0':s5_0(0)), gen_0':s5_0(+(1, n934_0)))) →IH
ack(gen_0':s5_0(0), *6_0)

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

### (18) Obligation:

TRS:
Rules:
numbersd(0')
d(x) → if(le(x, nr), x)
if(true, x) → cons(d(s(x)))
if(false, x) → nil
le(0', y) → true
le(s(x), 0') → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0')))))), 0')
ack(0', x) → s(x)
ack(s(x), 0') → ack(x, s(0'))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Types:
numbers :: cons:nil
d :: 0':s → cons:nil
0' :: 0':s
if :: true:false → 0':s → cons:nil
le :: 0':s → 0':s → true:false
nr :: 0':s
true :: true:false
cons :: cons:nil → cons:nil
s :: 0':s → 0':s
false :: true:false
nil :: cons:nil
ack :: 0':s → 0':s → 0':s
hole_cons:nil1_0 :: cons:nil
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_cons:nil4_0 :: Nat → cons:nil
gen_0':s5_0 :: Nat → 0':s

Lemmas:
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) → true, rt ∈ Ω(1 + n70)
ack(gen_0':s5_0(1), gen_0':s5_0(+(1, n934_0))) → *6_0, rt ∈ Ω(n9340)

Generator Equations:
gen_cons:nil4_0(0) ⇔ nil
gen_cons:nil4_0(+(x, 1)) ⇔ cons(gen_cons:nil4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))

No more defined symbols left to analyse.

### (19) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) → true, rt ∈ Ω(1 + n70)

### (21) Obligation:

TRS:
Rules:
numbersd(0')
d(x) → if(le(x, nr), x)
if(true, x) → cons(d(s(x)))
if(false, x) → nil
le(0', y) → true
le(s(x), 0') → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0')))))), 0')
ack(0', x) → s(x)
ack(s(x), 0') → ack(x, s(0'))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Types:
numbers :: cons:nil
d :: 0':s → cons:nil
0' :: 0':s
if :: true:false → 0':s → cons:nil
le :: 0':s → 0':s → true:false
nr :: 0':s
true :: true:false
cons :: cons:nil → cons:nil
s :: 0':s → 0':s
false :: true:false
nil :: cons:nil
ack :: 0':s → 0':s → 0':s
hole_cons:nil1_0 :: cons:nil
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_cons:nil4_0 :: Nat → cons:nil
gen_0':s5_0 :: Nat → 0':s

Lemmas:
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) → true, rt ∈ Ω(1 + n70)
ack(gen_0':s5_0(1), gen_0':s5_0(+(1, n934_0))) → *6_0, rt ∈ Ω(n9340)

Generator Equations:
gen_cons:nil4_0(0) ⇔ nil
gen_cons:nil4_0(+(x, 1)) ⇔ cons(gen_cons:nil4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))

No more defined symbols left to analyse.

### (22) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) → true, rt ∈ Ω(1 + n70)

### (24) Obligation:

TRS:
Rules:
numbersd(0')
d(x) → if(le(x, nr), x)
if(true, x) → cons(d(s(x)))
if(false, x) → nil
le(0', y) → true
le(s(x), 0') → false
le(s(x), s(y)) → le(x, y)
nrack(s(s(s(s(s(s(0')))))), 0')
ack(0', x) → s(x)
ack(s(x), 0') → ack(x, s(0'))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Types:
numbers :: cons:nil
d :: 0':s → cons:nil
0' :: 0':s
if :: true:false → 0':s → cons:nil
le :: 0':s → 0':s → true:false
nr :: 0':s
true :: true:false
cons :: cons:nil → cons:nil
s :: 0':s → 0':s
false :: true:false
nil :: cons:nil
ack :: 0':s → 0':s → 0':s
hole_cons:nil1_0 :: cons:nil
hole_0':s2_0 :: 0':s
hole_true:false3_0 :: true:false
gen_cons:nil4_0 :: Nat → cons:nil
gen_0':s5_0 :: Nat → 0':s

Lemmas:
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) → true, rt ∈ Ω(1 + n70)

Generator Equations:
gen_cons:nil4_0(0) ⇔ nil
gen_cons:nil4_0(+(x, 1)) ⇔ cons(gen_cons:nil4_0(x))
gen_0':s5_0(0) ⇔ 0'
gen_0':s5_0(+(x, 1)) ⇔ s(gen_0':s5_0(x))

No more defined symbols left to analyse.

### (25) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
le(gen_0':s5_0(n7_0), gen_0':s5_0(n7_0)) → true, rt ∈ Ω(1 + n70)