Termination proof of AProVE_24_purge

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 (⇒, 105 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 PisEmptyProof (⇔, 0 ms)

        ↳11 YES

    ↳12 RelADPP

        ↳13 RelADPCleverAfsProof (⇒, 103 ms)

        ↳14 QDP

        ↳15 MRRProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 DependencyGraphProof (⇔, 0 ms)

        ↳18 TRUE

    ↳19 RelADPP

        ↳20 RelADPReductionPairProof (⇔, 96 ms)

        ↳21 RelADPP

        ↳22 DAbsisEmptyProof (⇔, 0 ms)

        ↳23 YES

    ↳24 RelADPP

        ↳25 RelADPReductionPairProof (⇔, 98 ms)

        ↳26 RelADPP

        ↳27 DAbsisEmptyProof (⇔, 0 ms)

        ↳28 YES

    ↳29 RelADPP

        ↳30 RelADPReductionPairProof (⇔, 119 ms)

        ↳31 RelADPP

        ↳32 DAbsisEmptyProof (⇔, 0 ms)

        ↳33 YES

    ↳34 RelADPP

        ↳35 RelADPReductionPairProof (⇔, 95 ms)

        ↳36 RelADPP

        ↳37 DAbsisEmptyProof (⇔, 0 ms)

        ↳38 YES

(0) Obligation:

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

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
ifrm(true, n, cons(m, x)) → rm(n, x)
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
purge(nil) → nil
purge(cons(n, x)) → cons(n, purge(rm(n, x)))

The relative TRS consists of the following S rules:

consSwap(x, xs) → cons(x, xs)
purge(cons(x, xs)) → purge(consSwap(x, xs))
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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → EQ(x, y)
rm(n, nil) → nil
rm(n, cons(m, x)) → IFRM(eq(n, m), n, cons(m, x))
rm(n, cons(m, x)) → ifrm(EQ(n, m), n, cons(m, x))
ifrm(true, n, cons(m, x)) → RM(n, x)
ifrm(false, n, cons(m, x)) → cons(m, RM(n, x))
purge(nil) → nil
purge(cons(n, x)) → cons(n, PURGE(rm(n, x)))
purge(cons(n, x)) → cons(n, purge(RM(n, x)))

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
purge(cons(x, xs)) → PURGE(CONSSWAP(x, xs))
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,
3 Lassos,
Result: This relative DT problem is equivalent to 6 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

eq(s(x), s(y)) → EQ(x, y)

and relative ADPs:

ifrm(true, n, cons(m, x)) → rm(n, x)
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
purge(cons(x, xs)) → purge(consSwap(x, xs))
eq(0, 0) → true
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → 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:true = ifrm_3 = 0, 1 xs = rm_2 = 0, 1 false = nil = s_1 = eq_2 = 0 0 = purge_1 = cons_2 = 0, 1 consSwap_2 = 0 EQ_2 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
EQ(x1, x2) = EQ(x1, x2)
s(x1) = s(x1)
rm(x1, x2) = rm
cons(x1, x2) = cons
ifrm(x1, x2, x3) = x3
eq(x1, x2) = eq(x2)
purge(x1) = purge(x1)
xs = xs
consSwap(x1, x2) = x2
nil = nil
true = true
0 = 0
false = false

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

EQ₂ > [nil, true, 0, false]
s₁ > eq₁ > [nil, true, 0, false]
purge₁ > [rm, cons, xs] > [nil, true, 0, false]

Status:

EQ₂: [2,1]
s₁: multiset
rm: []
cons: []
eq₁: multiset
purge₁: multiset
xs: multiset
nil: multiset
true: multiset
0: multiset
false: multiset

(7) Obligation:

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

EQ0(s0(x), s0(y)) → EQ0(x, y)

The TRS R consists of the following rules:

ifrm(cons) → rm
consSwap(xs0) → cons
eq(s0(x)) → false0
eq(00) → false0
eq(s0(y)) → eq(y)
purge0(cons) → purge0(consSwap(xs0))
purge0(cons) → cons
consSwap(cons) → cons
rm → ifrm(cons)
purge0(nil0) → nil0
ifrm(cons) → cons
eq(00) → true0
rm → 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 dependency pairs:

EQ0(s0(x), s0(y)) → EQ0(x, y)

Strictly oriented rules of the TRS R:

eq(s0(x)) → false0
eq(00) → false0
eq(s0(y)) → eq(y)
purge0(cons) → cons
consSwap(cons) → cons
purge0(nil0) → nil0
ifrm(cons) → cons
eq(00) → true0
rm → nil0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2
POL(EQ0(x₁, x₂)) = x₁ + x₂
POL(cons) = 1
POL(consSwap(x₁)) = 1 + x₁
POL(eq(x₁)) = 2 + 2·x₁
POL(false0) = 1
POL(ifrm(x₁)) = 2·x₁
POL(nil0) = 1
POL(purge0(x₁)) = 2 + 2·x₁
POL(rm) = 2
POL(s0(x₁)) = 1 + x₁
POL(true0) = 1
POL(xs0) = 0

(9) Obligation:

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

ifrm(cons) → rm
consSwap(xs0) → cons
purge0(cons) → purge0(consSwap(xs0))
rm → ifrm(cons)

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:

ifrm(false, n, cons(m, x)) → cons(m, RM(n, x))
ifrm(true, n, cons(m, x)) → RM(n, x)
rm(n, cons(m, x)) → IFRM(eq(n, m), n, cons(m, x))

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(x, xs)) → purge(consSwap(x, xs))
eq(0, 0) → true
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → 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:true = ifrm_3 = 0, 1 xs = rm_2 = 0 IFRM_3 = 0, 1 false = nil = s_1 = eq_2 = 0, 1 RM_2 = 0 0 = purge_1 = cons_2 = 0 consSwap_2 = 0
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
RM(x1, x2) = RM(x2)
cons(x1, x2) = cons(x2)
IFRM(x1, x2, x3) = x3
eq(x1, x2) = eq
false = false
true = true
0 = 0
s(x1) = x1
purge(x1) = purge(x1)
rm(x1, x2) = x2
ifrm(x1, x2, x3) = x3
nil = nil
xs = xs
consSwap(x1, x2) = consSwap(x2)

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

[RM₁, cons₁, purge₁, consSwap₁] > [eq, false] > true
[RM₁, cons₁, purge₁, consSwap₁] > nil
[RM₁, cons₁, purge₁, consSwap₁] > xs

Status:

RM₁: [1]
cons₁: [1]
eq: multiset
false: multiset
true: multiset
0: multiset
purge₁: [1]
nil: multiset
xs: multiset
consSwap₁: [1]

(14) Obligation:

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

RM(cons(x)) → IFRM(cons(x))
IFRM(cons(x)) → RM(x)

The TRS R consists of the following rules:

ifrm(cons(x)) → rm(x)
consSwap(xs0) → cons(xs0)
eq → false0
eq → eq
purge0(cons(xs0)) → purge0(consSwap(xs0))
purge0(cons(x)) → cons(purge0(rm(x)))
consSwap(cons(xs0)) → cons(consSwap(xs0))
rm(cons(x)) → ifrm(cons(x))
purge0(nil0) → nil0
ifrm(cons(x)) → cons(rm(x))
eq → true0
rm(nil0) → nil0

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

(15) MRRProof (EQUIVALENT transformation)

IFRM(cons(x)) → RM(x)

Strictly oriented rules of the TRS R:

ifrm(cons(x)) → rm(x)
eq → false0
eq → true0

Used ordering: Polynomial interpretation [POLO]:

POL(IFRM(x₁)) = x₁
POL(RM(x₁)) = x₁
POL(cons(x₁)) = 1 + 2·x₁
POL(consSwap(x₁)) = 1 + 2·x₁
POL(eq) = 2
POL(false0) = 1
POL(ifrm(x₁)) = x₁
POL(nil0) = 0
POL(purge0(x₁)) = x₁
POL(rm(x₁)) = x₁
POL(true0) = 0
POL(xs0) = 1

(16) Obligation:

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

RM(cons(x)) → IFRM(cons(x))

The TRS R consists of the following rules:

consSwap(xs0) → cons(xs0)
eq → eq
purge0(cons(xs0)) → purge0(consSwap(xs0))
purge0(cons(x)) → cons(purge0(rm(x)))
consSwap(cons(xs0)) → cons(consSwap(xs0))
rm(cons(x)) → ifrm(cons(x))
purge0(nil0) → nil0
ifrm(cons(x)) → cons(rm(x))
rm(nil0) → nil0

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(18) TRUE

(19) Obligation:

Relative ADP Problem with
absolute ADPs:

purge(cons(n, x)) → cons(n, PURGE(rm(n, x)))

and relative ADPs:

ifrm(true, n, cons(m, x)) → rm(n, x)
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(x, xs)) → PURGE(CONSSWAP(x, xs))
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
eq(0, 0) → true
rm(n, nil) → 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:

purge(cons(n, x)) → cons(n, PURGE(rm(n, x)))

Relative ADPs:

ifrm(true, n, cons(m, x)) → rm(n, x)
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
eq(0, 0) → true
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 1
POL(CONSSWAP(x₁, x₂)) = 0
POL(EQ(x₁, x₂)) = 2 + 2·x₁·x₂
POL(IFRM(x₁, x₂, x₃)) = 2 + x₁ + 2·x₂
POL(PURGE(x₁)) = 2 + 3·x₁
POL(RM(x₁, x₂)) = 2 + 2·x₂
POL(cons(x₁, x₂)) = 3 + 2·x₁ + x₂
POL(consSwap(x₁, x₂)) = 3 + 2·x₁ + x₂
POL(eq(x₁, x₂)) = 0
POL(false) = 0
POL(ifrm(x₁, x₂, x₃)) = x₁ + x₃
POL(nil) = 0
POL(purge(x₁)) = x₁
POL(rm(x₁, x₂)) = x₂
POL(s(x₁)) = 3 + 3·x₁
POL(true) = 0
POL(xs) = 2

(21) Obligation:

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

ifrm(true, n, cons(m, x)) → rm(n, x)
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(x, xs)) → PURGE(CONSSWAP(x, xs))
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
eq(0, 0) → true
rm(n, nil) → 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:

purge(cons(n, x)) → cons(n, purge(RM(n, x)))

and relative ADPs:

ifrm(true, n, cons(m, x)) → rm(n, x)
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(x, xs)) → PURGE(CONSSWAP(x, xs))
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
eq(0, 0) → true
rm(n, nil) → 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:

purge(cons(n, x)) → cons(n, purge(RM(n, x)))

Relative ADPs:

ifrm(true, n, cons(m, x)) → rm(n, x)
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
eq(0, 0) → true
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 3
POL(CONSSWAP(x₁, x₂)) = 0
POL(EQ(x₁, x₂)) = 2·x₁·x₂
POL(IFRM(x₁, x₂, x₃)) = 2 + 2·x₂
POL(PURGE(x₁)) = 2 + 3·x₁
POL(RM(x₁, x₂)) = 1
POL(cons(x₁, x₂)) = 2
POL(consSwap(x₁, x₂)) = 1 + x₂
POL(eq(x₁, x₂)) = 3
POL(false) = 2
POL(ifrm(x₁, x₂, x₃)) = x₁ + 3·x₂
POL(nil) = 0
POL(purge(x₁)) = 2 + 3·x₁
POL(rm(x₁, x₂)) = 3 + 3·x₁
POL(s(x₁)) = 3 + 3·x₁
POL(true) = 3
POL(xs) = 1

(26) Obligation:

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

ifrm(true, n, cons(m, x)) → rm(n, x)
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(x, xs)) → PURGE(CONSSWAP(x, xs))
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
eq(0, 0) → true
rm(n, nil) → 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:

purge(nil) → nil

and relative ADPs:

ifrm(true, n, cons(m, x)) → rm(n, x)
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
eq(0, 0) → true
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(cons(x, xs)) → PURGE(CONSSWAP(x, xs))

(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:

purge(nil) → nil

Relative ADPs:

ifrm(true, n, cons(m, x)) → rm(n, x)
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
eq(0, 0) → true
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 2
POL(CONSSWAP(x₁, x₂)) = 0
POL(EQ(x₁, x₂)) = 2·x₁·x₂
POL(IFRM(x₁, x₂, x₃)) = 2 + 2·x₁ + 2·x₂
POL(PURGE(x₁)) = 1 + x₁
POL(RM(x₁, x₂)) = 2 + 2·x₁
POL(cons(x₁, x₂)) = 1 + 3·x₂
POL(consSwap(x₁, x₂)) = 1 + 3·x₂
POL(eq(x₁, x₂)) = 0
POL(false) = 0
POL(ifrm(x₁, x₂, x₃)) = 3·x₁ + x₃
POL(nil) = 0
POL(purge(x₁)) = 1 + 3·x₁
POL(rm(x₁, x₂)) = x₂
POL(s(x₁)) = 1 + 3·x₁
POL(true) = 0
POL(xs) = 2

(31) Obligation:

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

ifrm(true, n, cons(m, x)) → rm(n, x)
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(x, xs)) → PURGE(CONSSWAP(x, xs))
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
eq(0, 0) → true
rm(n, nil) → 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:

purge(cons(n, x)) → cons(n, PURGE(rm(n, x)))

and relative ADPs:

ifrm(true, n, cons(m, x)) → rm(n, x)
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(x, xs)) → PURGE(CONSSWAP(x, xs))
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
eq(0, 0) → true
rm(n, nil) → 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:

purge(cons(n, x)) → cons(n, PURGE(rm(n, x)))

Relative ADPs:

ifrm(true, n, cons(m, x)) → rm(n, x)
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
eq(0, 0) → true
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 1
POL(CONSSWAP(x₁, x₂)) = 0
POL(EQ(x₁, x₂)) = 2 + 2·x₁·x₂
POL(IFRM(x₁, x₂, x₃)) = 2 + x₁ + 2·x₂
POL(PURGE(x₁)) = 2 + 3·x₁
POL(RM(x₁, x₂)) = 2 + 2·x₂
POL(cons(x₁, x₂)) = 3 + 2·x₁ + x₂
POL(consSwap(x₁, x₂)) = 3 + 2·x₁ + x₂
POL(eq(x₁, x₂)) = 0
POL(false) = 0
POL(ifrm(x₁, x₂, x₃)) = x₁ + x₃
POL(nil) = 0
POL(purge(x₁)) = x₁
POL(rm(x₁, x₂)) = x₂
POL(s(x₁)) = 3 + 3·x₁
POL(true) = 0
POL(xs) = 2

(36) Obligation:

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

ifrm(true, n, cons(m, x)) → rm(n, x)
consSwap(x, xs) → cons(x, xs)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
purge(cons(n, x)) → cons(n, purge(rm(n, x)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(x, xs)) → PURGE(CONSSWAP(x, xs))
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
eq(0, 0) → true
rm(n, nil) → nil

(37) 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) PisEmptyProof (EQUIVALENT transformation)

(11) YES

(12) Obligation:

(13) RelADPCleverAfsProof (SOUND transformation)

(14) Obligation:

(15) MRRProof (EQUIVALENT transformation)

(16) Obligation:

(17) DependencyGraphProof (EQUIVALENT transformation)

(18) TRUE

(19) Obligation:

(20) RelADPReductionPairProof (EQUIVALENT transformation)

(21) Obligation:

(22) DAbsisEmptyProof (EQUIVALENT transformation)

(23) YES

(24) Obligation:

(25) RelADPReductionPairProof (EQUIVALENT transformation)

(26) Obligation:

(27) DAbsisEmptyProof (EQUIVALENT transformation)

(28) YES

(29) Obligation:

(30) RelADPReductionPairProof (EQUIVALENT transformation)

(31) Obligation:

(32) DAbsisEmptyProof (EQUIVALENT transformation)

(33) YES

(34) Obligation:

(35) RelADPReductionPairProof (EQUIVALENT transformation)

(36) Obligation:

(37) DAbsisEmptyProof (EQUIVALENT transformation)

(38) YES