Termination proof of AProVE_24

Termination of the given RelTRS could be proven:

0 RelTRS

↳1 RelTRStoRelADPProof (⇔, 0 ms)

↳2 RelADPP

↳3 RelADPDepGraphProof (⇔, 0 ms)

↳4 AND

    ↳5 RelADPP

        ↳6 RelADPCleverAfsProof (⇒, 112 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 MRRProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 PisEmptyProof (⇔, 0 ms)

        ↳13 YES

    ↳14 RelADPP

        ↳15 RelADPCleverAfsProof (⇒, 101 ms)

        ↳16 QDP

        ↳17 MRRProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 MRRProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 PisEmptyProof (⇔, 0 ms)

        ↳22 YES

    ↳23 RelADPP

        ↳24 RelADPReductionPairProof (⇔, 60 ms)

        ↳25 RelADPP

        ↳26 DAbsisEmptyProof (⇔, 0 ms)

        ↳27 YES

(0) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

remove(nil, x) → nil
remove(cons(x, xs), 0) → cons(x, xs)
remove(cons(x, xs), s(y)) → remove(xs, y)
first(nil, x) → nil
first(cons(x, xs), 0) → nil
first(cons(x, xs), s(y)) → cons(x, first(xs, y))
sublist(nil, x) → nil
sublist(cons(x, xs), 0) → nil
sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), sublist(remove(xs, y), s(y)))

The relative TRS consists of the following S rules:

sublist(cons(x, cons(y, xs)), z) → sublist(cons(y, cons(x, xs)), z)

(1) RelTRStoRelADPProof (EQUIVALENT transformation)

We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].

(2) Obligation:

Relative ADP Problem with
absolute ADPs:

remove(nil, x) → nil
remove(cons(x, xs), 0) → cons(x, xs)
remove(cons(x, xs), s(y)) → REMOVE(xs, y)
first(nil, x) → nil
first(cons(x, xs), 0) → nil
first(cons(x, xs), s(y)) → cons(x, FIRST(xs, y))
sublist(nil, x) → nil
sublist(cons(x, xs), 0) → nil
sublist(cons(x, xs), s(y)) → cons(cons(x, FIRST(xs, y)), sublist(remove(xs, y), s(y)))
sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), SUBLIST(remove(xs, y), s(y)))
sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), sublist(REMOVE(xs, y), s(y)))

and relative ADPs:

sublist(cons(x, cons(y, xs)), z) → SUBLIST(cons(y, cons(x, xs)), z)

(3) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
3 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 3 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

first(cons(x, xs), s(y)) → cons(x, FIRST(xs, y))

and relative ADPs:

remove(cons(x, xs), 0) → cons(x, xs)
sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), sublist(remove(xs, y), s(y)))
first(nil, x) → nil
remove(nil, x) → nil
sublist(nil, x) → nil
sublist(cons(x, xs), 0) → nil
sublist(cons(x, cons(y, xs)), z) → sublist(cons(y, cons(x, xs)), z)
remove(cons(x, xs), s(y)) → remove(xs, y)
first(cons(x, xs), 0) → nil

(6) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 = FIRST_2 = 0 = cons_2 = 0 remove_2 = 1 first_2 = 0 sublist_2 = 0 nil =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
FIRST(x1, x2) = FIRST(x1, x2)
cons(x1, x2) = x2
s(x1) = s(x1)
sublist(x1, x2) = x2
first(x1, x2) = first(x2)
remove(x1, x2) = remove(x1)
0 = 0
nil = nil

Recursive path order with status [RPO].
Quasi-Precedence:

s₁ > FIRST₂ > nil
first₁ > nil
remove₁ > nil
0 > nil

Status:

FIRST₂: multiset
s₁: multiset
first₁: [1]
remove₁: [1]
0: multiset
nil: multiset

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FIRST0(cons(xs), s0(y)) → FIRST0(xs, y)

The TRS R consists of the following rules:

remove(cons(xs)) → cons(xs)
sublist(s0(y)) → cons(sublist(s0(y)))
first(x) → nil0
remove(nil0) → nil0
sublist(x) → nil0
sublist(00) → nil0
sublist(z) → sublist(z)
first(s0(y)) → cons(first(y))
remove(cons(xs)) → remove(xs)
first(00) → nil0

Q is empty.
We have to consider all (P,Q,R)-chains.

(8) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

remove(cons(xs)) → cons(xs)
first(x) → nil0
remove(nil0) → nil0
sublist(x) → nil0
sublist(00) → nil0
first(00) → nil0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(FIRST0(x₁, x₂)) = x₁ + x₂
POL(cons(x₁)) = x₁
POL(first(x₁)) = 2 + 2·x₁
POL(nil0) = 0
POL(remove(x₁)) = 1 + 2·x₁
POL(s0(x₁)) = x₁
POL(sublist(x₁)) = 1 + 2·x₁

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FIRST0(cons(xs), s0(y)) → FIRST0(xs, y)

The TRS R consists of the following rules:

sublist(s0(y)) → cons(sublist(s0(y)))
sublist(z) → sublist(z)
first(s0(y)) → cons(first(y))
remove(cons(xs)) → remove(xs)

Q is empty.
We have to consider all (P,Q,R)-chains.

(10) MRRProof (EQUIVALENT transformation)

FIRST0(cons(xs), s0(y)) → FIRST0(xs, y)

Strictly oriented rules of the TRS R:

first(s0(y)) → cons(first(y))

Used ordering: Polynomial interpretation [POLO]:

POL(FIRST0(x₁, x₂)) = x₁ + x₂
POL(cons(x₁)) = x₁
POL(first(x₁)) = 1 + x₁
POL(remove(x₁)) = x₁
POL(s0(x₁)) = 1 + 2·x₁
POL(sublist(x₁)) = 1 + 2·x₁

(11) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sublist(s0(y)) → cons(sublist(s0(y)))
sublist(z) → sublist(z)
remove(cons(xs)) → remove(xs)

Q is empty.
We have to consider all (P,Q,R)-chains.

(12) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(13) YES

(14) Obligation:

Relative ADP Problem with
absolute ADPs:

remove(cons(x, xs), s(y)) → REMOVE(xs, y)

and relative ADPs:

remove(cons(x, xs), 0) → cons(x, xs)
sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), sublist(remove(xs, y), s(y)))
first(nil, x) → nil
remove(nil, x) → nil
sublist(nil, x) → nil
sublist(cons(x, xs), 0) → nil
sublist(cons(x, cons(y, xs)), z) → sublist(cons(y, cons(x, xs)), z)
first(cons(x, xs), s(y)) → cons(x, first(xs, y))
first(cons(x, xs), 0) → nil

(15) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 = REMOVE_2 = 0 = remove_2 = 1 cons_2 = 0 first_2 = 0 sublist_2 = 0 nil =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
REMOVE(x1, x2) = REMOVE(x1, x2)
cons(x1, x2) = x2
s(x1) = s(x1)
sublist(x1, x2) = x2
first(x1, x2) = first(x2)
remove(x1, x2) = remove(x1)
0 = 0
nil = nil

Recursive path order with status [RPO].
Quasi-Precedence:

s₁ > REMOVE₂ > nil
first₁ > nil
remove₁ > nil
0 > nil

Status:

REMOVE₂: multiset
s₁: multiset
first₁: [1]
remove₁: [1]
0: multiset
nil: multiset

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

REMOVE0(cons(xs), s0(y)) → REMOVE0(xs, y)

The TRS R consists of the following rules:

remove(cons(xs)) → cons(xs)
sublist(s0(y)) → cons(sublist(s0(y)))
first(x) → nil0
remove(nil0) → nil0
sublist(x) → nil0
sublist(00) → nil0
sublist(z) → sublist(z)
first(s0(y)) → cons(first(y))
remove(cons(xs)) → remove(xs)
first(00) → nil0

Q is empty.
We have to consider all (P,Q,R)-chains.

(17) MRRProof (EQUIVALENT transformation)

remove(cons(xs)) → cons(xs)
first(x) → nil0
remove(nil0) → nil0
sublist(x) → nil0
sublist(00) → nil0
first(00) → nil0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(REMOVE0(x₁, x₂)) = x₁ + x₂
POL(cons(x₁)) = x₁
POL(first(x₁)) = 2 + 2·x₁
POL(nil0) = 0
POL(remove(x₁)) = 1 + 2·x₁
POL(s0(x₁)) = x₁
POL(sublist(x₁)) = 1 + 2·x₁

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

REMOVE0(cons(xs), s0(y)) → REMOVE0(xs, y)

The TRS R consists of the following rules:

sublist(s0(y)) → cons(sublist(s0(y)))
sublist(z) → sublist(z)
first(s0(y)) → cons(first(y))
remove(cons(xs)) → remove(xs)

Q is empty.
We have to consider all (P,Q,R)-chains.

(19) MRRProof (EQUIVALENT transformation)

REMOVE0(cons(xs), s0(y)) → REMOVE0(xs, y)

Strictly oriented rules of the TRS R:

first(s0(y)) → cons(first(y))

Used ordering: Polynomial interpretation [POLO]:

POL(REMOVE0(x₁, x₂)) = x₁ + x₂
POL(cons(x₁)) = x₁
POL(first(x₁)) = 1 + x₁
POL(remove(x₁)) = x₁
POL(s0(x₁)) = 1 + 2·x₁
POL(sublist(x₁)) = 1 + 2·x₁

(20) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sublist(s0(y)) → cons(sublist(s0(y)))
sublist(z) → sublist(z)
remove(cons(xs)) → remove(xs)

Q is empty.
We have to consider all (P,Q,R)-chains.

(21) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(22) YES

(23) Obligation:

Relative ADP Problem with
absolute ADPs:

sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), SUBLIST(remove(xs, y), s(y)))

and relative ADPs:

remove(cons(x, xs), 0) → cons(x, xs)
sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), sublist(remove(xs, y), s(y)))
first(nil, x) → nil
remove(nil, x) → nil
sublist(nil, x) → nil
sublist(cons(x, xs), 0) → nil
sublist(cons(x, cons(y, xs)), z) → SUBLIST(cons(y, cons(x, xs)), z)
first(cons(x, xs), s(y)) → cons(x, first(xs, y))
remove(cons(x, xs), s(y)) → remove(xs, y)
first(cons(x, xs), 0) → nil

(24) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:

sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), SUBLIST(remove(xs, y), s(y)))

Relative ADPs:

remove(cons(x, xs), 0) → cons(x, xs)
sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), sublist(remove(xs, y), s(y)))
first(nil, x) → nil
remove(nil, x) → nil
sublist(nil, x) → nil
sublist(cons(x, xs), 0) → nil
first(cons(x, xs), s(y)) → cons(x, first(xs, y))
remove(cons(x, xs), s(y)) → remove(xs, y)
first(cons(x, xs), 0) → nil

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0
POL(FIRST(x₁, x₂)) = 2
POL(REMOVE(x₁, x₂)) = 2
POL(SUBLIST(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(cons(x₁, x₂)) = 3 + 2·x₂
POL(first(x₁, x₂)) = x₁
POL(nil) = 0
POL(remove(x₁, x₂)) = x₁
POL(s(x₁)) = 3
POL(sublist(x₁, x₂)) = 3 + 3·x₁

(25) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

sublist(cons(x, xs), s(y)) → cons(cons(x, first(xs, y)), sublist(remove(xs, y), s(y)))
remove(cons(x, xs), 0) → cons(x, xs)
first(nil, x) → nil
remove(nil, x) → nil
sublist(nil, x) → nil
sublist(cons(x, xs), 0) → nil
sublist(cons(x, cons(y, xs)), z) → SUBLIST(cons(y, cons(x, xs)), z)
first(cons(x, xs), s(y)) → cons(x, first(xs, y))
remove(cons(x, xs), s(y)) → remove(xs, y)
first(cons(x, xs), 0) → nil

(26) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(0) Obligation:

(1) RelTRStoRelADPProof (EQUIVALENT transformation)

(2) Obligation:

(3) RelADPDepGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) RelADPCleverAfsProof (SOUND transformation)

(7) Obligation:

(8) MRRProof (EQUIVALENT transformation)

(9) Obligation:

(10) MRRProof (EQUIVALENT transformation)

(11) Obligation:

(12) PisEmptyProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) RelADPCleverAfsProof (SOUND transformation)

(16) Obligation:

(17) MRRProof (EQUIVALENT transformation)

(18) Obligation:

(19) MRRProof (EQUIVALENT transformation)

(20) Obligation:

(21) PisEmptyProof (EQUIVALENT transformation)

(22) YES

(23) Obligation:

(24) RelADPReductionPairProof (EQUIVALENT transformation)

(25) Obligation:

(26) DAbsisEmptyProof (EQUIVALENT transformation)

(27) YES