YES Termination proof of AProVE_24_subsets_2.ari

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 (⇒, 114 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 12 ms)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔, 0 ms)
        ↳11 YES
    ↳12 RelADPP
        ↳13 RelADPCleverAfsProof (⇒, 122 ms)
        ↳14 QDP
        ↳15 MRRProof (⇔, 14 ms)
        ↳16 QDP
        ↳17 PisEmptyProof (⇔, 0 ms)
        ↳18 YES
    ↳19 RelADPP
        ↳20 RelADPReductionPairProof (⇔, 110 ms)
        ↳21 RelADPP
        ↳22 DAbsisEmptyProof (⇔, 0 ms)
        ↳23 YES
    ↳24 RelADPP
        ↳25 RelADPReductionPairProof (⇔, 108 ms)
        ↳26 RelADPP
        ↳27 DAbsisEmptyProof (⇔, 0 ms)
        ↳28 YES
    ↳29 RelADPP
        ↳30 RelADPReductionPairProof (⇔, 125 ms)
        ↳31 RelADPP
        ↳32 DAbsisEmptyProof (⇔, 0 ms)
        ↳33 YES
    ↳34 RelADPP
        ↳35 RelADPReductionPairProof (⇔, 108 ms)
        ↳36 RelADPP
        ↳37 DAbsisEmptyProof (⇔, 0 ms)
        ↳38 YES
    ↳39 RelADPP
        ↳40 RelADPReductionPairProof (⇔, 101 ms)
        ↳41 RelADPP
        ↳42 DAbsisEmptyProof (⇔, 0 ms)
        ↳43 YES
    ↳44 RelADPP
        ↳45 RelADPReductionPairProof (⇔, 111 ms)
        ↳46 RelADPP
        ↳47 DAbsisEmptyProof (⇔, 0 ms)
        ↳48 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:

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

(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:

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

(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,
5 Lassos,
Result: This relative DT problem is equivalent to 8 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
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, xs), z) → sublist(consSwap(x, xs), z)
first(cons(x, xs), s(y)) → cons(x, first(xs, y))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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 = REMOVE_2 = 0 = remove_2 = 1 cons_2 = 0 first_2 = 0, 1 sublist_2 = 0 consSwap_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
remove(x1, x2)  =  remove(x1)
0  =  0
nil  =  nil
consSwap(x1, x2)  =  x2

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

REMOVE2 > [remove1, nil]
[s1, first] > [remove1, nil]
0 > [remove1, nil]

Status:
REMOVE2: multiset
s1: multiset
first: []
remove1: [1]
0: multiset
nil: multiset

(7) 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:

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

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 dependency pairs:

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

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

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

(9) 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)
first → cons(first)
consSwap(cons(xs)) → cons(consSwap(xs))
remove(cons(xs)) → remove(xs)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) YES

(12) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

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

(13) 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, 1 sublist_2 = 0 consSwap_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
remove(x1, x2)  =  remove(x1)
0  =  0
nil  =  nil
consSwap(x1, x2)  =  x2

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

FIRST2 > [remove1, nil]
[s1, first] > [remove1, nil]
0 > [remove1, nil]

Status:
FIRST2: multiset
s1: multiset
first: []
remove1: [1]
0: multiset
nil: multiset

(14) 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:

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

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

(15) 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 dependency pairs:

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

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

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

(16) 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)
first → cons(first)
consSwap(cons(xs)) → cons(consSwap(xs))
remove(cons(xs)) → remove(xs)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) YES

(19) 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:

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

(20) 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:

consSwap(x, xs) → cons(x, xs)
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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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(CONSSWAP(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 2 + x1·x2 + x2   
POL(REMOVE(x1, x2)) = 2   
POL(SUBLIST(x1, x2)) = 2 + 3·x1   
POL(cons(x1, x2)) = 1 + x2   
POL(consSwap(x1, x2)) = 1 + x2   
POL(first(x1, x2)) = x1   
POL(nil) = 0   
POL(remove(x1, x2)) = x1   
POL(s(x1)) = 3   
POL(sublist(x1, x2)) = 2·x1 + 3·x2   

(21) Obligation:

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

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

(22) DAbsisEmptyProof (EQUIVALENT transformation)

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

(23) YES

(24) Obligation:

Relative ADP Problem with
absolute ADPs:

sublist(cons(x, xs), 0) → nil

and relative ADPs:

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

(25) 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), 0) → nil

Relative ADPs:

consSwap(x, xs) → cons(x, xs)
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
first(cons(x, xs), s(y)) → cons(x, first(xs, y))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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(CONSSWAP(x1, x2)) = 0   
POL(FIRST(x1, x2)) = x2   
POL(REMOVE(x1, x2)) = 2   
POL(SUBLIST(x1, x2)) = 2·x1   
POL(cons(x1, x2)) = 1   
POL(consSwap(x1, x2)) = 1   
POL(first(x1, x2)) = 3·x2   
POL(nil) = 0   
POL(remove(x1, x2)) = 1   
POL(s(x1)) = 2 + 2·x1   
POL(sublist(x1, x2)) = 2 + 2·x1   

(26) Obligation:

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

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

(27) DAbsisEmptyProof (EQUIVALENT transformation)

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

(28) YES

(29) Obligation:

Relative ADP Problem with
absolute ADPs:

sublist(nil, x) → nil

and relative ADPs:

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

(30) 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(nil, x) → nil

Relative ADPs:

consSwap(x, xs) → cons(x, xs)
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(cons(x, xs), 0) → nil
first(cons(x, xs), s(y)) → cons(x, first(xs, y))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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(CONSSWAP(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 2 + 2·x1   
POL(REMOVE(x1, x2)) = 2   
POL(SUBLIST(x1, x2)) = 1 + 2·x1   
POL(cons(x1, x2)) = 1   
POL(consSwap(x1, x2)) = 1   
POL(first(x1, x2)) = 2·x2   
POL(nil) = 0   
POL(remove(x1, x2)) = 1   
POL(s(x1)) = 3 + x1   
POL(sublist(x1, x2)) = 3·x2   

(31) Obligation:

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

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

(32) DAbsisEmptyProof (EQUIVALENT transformation)

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

(33) YES

(34) 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:

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

(35) 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:

consSwap(x, xs) → cons(x, xs)
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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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(CONSSWAP(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 2   
POL(REMOVE(x1, x2)) = 1 + 2·x1   
POL(SUBLIST(x1, x2)) = 2 + 3·x1 + 3·x2   
POL(cons(x1, x2)) = 1 + 3·x2   
POL(consSwap(x1, x2)) = 1 + 3·x2   
POL(first(x1, x2)) = x1   
POL(nil) = 0   
POL(remove(x1, x2)) = x1   
POL(s(x1)) = 3   
POL(sublist(x1, x2)) = x1   

(36) Obligation:

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

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

(37) DAbsisEmptyProof (EQUIVALENT transformation)

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

(38) YES

(39) 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:

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

(40) 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:

consSwap(x, xs) → cons(x, xs)
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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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(CONSSWAP(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 2 + x1·x2 + x2   
POL(REMOVE(x1, x2)) = 2   
POL(SUBLIST(x1, x2)) = 2 + 3·x1   
POL(cons(x1, x2)) = 1 + x2   
POL(consSwap(x1, x2)) = 1 + x2   
POL(first(x1, x2)) = x1   
POL(nil) = 0   
POL(remove(x1, x2)) = x1   
POL(s(x1)) = 3   
POL(sublist(x1, x2)) = 2·x1 + 3·x2   

(41) Obligation:

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

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

(42) DAbsisEmptyProof (EQUIVALENT transformation)

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

(43) YES

(44) 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:

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

(45) 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:

consSwap(x, xs) → cons(x, xs)
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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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) = 1   
POL(CONSSWAP(x1, x2)) = 0   
POL(FIRST(x1, x2)) = 1   
POL(REMOVE(x1, x2)) = x2   
POL(SUBLIST(x1, x2)) = 1 + 2·x1 + 3·x2   
POL(cons(x1, x2)) = 1   
POL(consSwap(x1, x2)) = 1   
POL(first(x1, x2)) = x1   
POL(nil) = 1   
POL(remove(x1, x2)) = 1   
POL(s(x1)) = 3 + 2·x1   
POL(sublist(x1, x2)) = 2·x1 + 2·x2   

(46) Obligation:

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

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

(47) DAbsisEmptyProof (EQUIVALENT transformation)

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

(48) YES