(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
numbers → d(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)
nr → ack(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 [ ].
(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:
numbers → d(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)
nr → ack(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:
numbers → d(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)
nr → ack(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
(7) TypeInferenceProof (BOTH BOUNDS(ID, ID) transformation)
Infered types.
(8) Obligation:
TRS:
Rules:
numbers → d(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)
nr → ack(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,
ackThey will be analysed ascendingly in the following order:
le < d
(10) Obligation:
TRS:
Rules:
numbers →
d(
0')
d(
x) →
if(
le(
x,
nr),
x)
if(
true,
x) →
cons(
d(
s(
x)))
if(
false,
x) →
nille(
0',
y) →
truele(
s(
x),
0') →
falsele(
s(
x),
s(
y)) →
le(
x,
y)
nr →
ack(
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 + n7
0)
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).
(12) Complex Obligation (BEST)
(13) Obligation:
TRS:
Rules:
numbers →
d(
0')
d(
x) →
if(
le(
x,
nr),
x)
if(
true,
x) →
cons(
d(
s(
x)))
if(
false,
x) →
nille(
0',
y) →
truele(
s(
x),
0') →
falsele(
s(
x),
s(
y)) →
le(
x,
y)
nr →
ack(
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:
numbers →
d(
0')
d(
x) →
if(
le(
x,
nr),
x)
if(
true,
x) →
cons(
d(
s(
x)))
if(
false,
x) →
nille(
0',
y) →
truele(
s(
x),
0') →
falsele(
s(
x),
s(
y)) →
le(
x,
y)
nr →
ack(
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(n934
0)
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).
(17) Complex Obligation (BEST)
(18) Obligation:
TRS:
Rules:
numbers →
d(
0')
d(
x) →
if(
le(
x,
nr),
x)
if(
true,
x) →
cons(
d(
s(
x)))
if(
false,
x) →
nille(
0',
y) →
truele(
s(
x),
0') →
falsele(
s(
x),
s(
y)) →
le(
x,
y)
nr →
ack(
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)
(20) BOUNDS(n^1, INF)
(21) Obligation:
TRS:
Rules:
numbers →
d(
0')
d(
x) →
if(
le(
x,
nr),
x)
if(
true,
x) →
cons(
d(
s(
x)))
if(
false,
x) →
nille(
0',
y) →
truele(
s(
x),
0') →
falsele(
s(
x),
s(
y)) →
le(
x,
y)
nr →
ack(
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)
(23) BOUNDS(n^1, INF)
(24) Obligation:
TRS:
Rules:
numbers →
d(
0')
d(
x) →
if(
le(
x,
nr),
x)
if(
true,
x) →
cons(
d(
s(
x)))
if(
false,
x) →
nille(
0',
y) →
truele(
s(
x),
0') →
falsele(
s(
x),
s(
y)) →
le(
x,
y)
nr →
ack(
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)
(26) BOUNDS(n^1, INF)