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 (⇒, 53 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 (⇒, 48 ms)

        ↳16 QDP

        ↳17 MRRProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 DependencyGraphProof (⇔, 0 ms)

        ↳20 TRUE

    ↳21 RelADPP

        ↳22 RelADPReductionPairProof (⇔, 56 ms)

        ↳23 RelADPP

        ↳24 DAbsisEmptyProof (⇔, 0 ms)

        ↳25 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:

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

(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:

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

(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:

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

(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 = true = eq_2 = 0, 1 ifrm_3 = 0, 1 0 = purge_1 = 0 cons_2 = 0 rm_2 = 0 EQ_2 = false = 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.
EQ(x1, x2) = EQ(x1, x2)
s(x1) = s(x1)
purge(x1) = purge
cons(x1, x2) = x2
rm(x1, x2) = x2
ifrm(x1, x2, x3) = x3
eq(x1, x2) = eq
true = true
false = false
0 = 0
nil = nil

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

purge > nil
eq > [s₁, false]
eq > [true, 0]

Status:

EQ₂: multiset
s₁: multiset
purge: []
eq: []
true: multiset
false: multiset
0: multiset
nil: 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(x)) → rm(x)
ifrm(cons(x)) → cons(rm(x))
eq → false0
eq → eq
eq → true0
purge → cons(purge)
rm(nil0) → nil0
rm(cons(x)) → ifrm(cons(x))
purge → nil0
purge → purge

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:

eq → false0
eq → true0

Used ordering: Polynomial interpretation [POLO]:

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

(9) 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(x)) → rm(x)
ifrm(cons(x)) → cons(rm(x))
eq → eq
purge → cons(purge)
rm(nil0) → nil0
rm(cons(x)) → ifrm(cons(x))
purge → nil0
purge → purge

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

(10) MRRProof (EQUIVALENT transformation)

EQ0(s0(x), s0(y)) → EQ0(x, y)

Strictly oriented rules of the TRS R:

rm(nil0) → nil0
purge → nil0

Used ordering: Polynomial interpretation [POLO]:

POL(EQ0(x₁, x₂)) = x₁ + x₂
POL(cons(x₁)) = x₁
POL(eq) = 0
POL(ifrm(x₁)) = 2 + 2·x₁
POL(nil0) = 1
POL(purge) = 2
POL(rm(x₁)) = 2 + 2·x₁
POL(s0(x₁)) = 1 + x₁

(11) Obligation:

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

ifrm(cons(x)) → rm(x)
ifrm(cons(x)) → cons(rm(x))
eq → eq
purge → cons(purge)
rm(cons(x)) → ifrm(cons(x))
purge → purge

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:

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:

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)))
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(n, cons(m, x))) → purge(cons(m, cons(n, x)))

(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 = true = eq_2 = 0, 1 RM_2 = 0 ifrm_3 = 0, 1 0 = purge_1 = cons_2 = 0 rm_2 = 0 IFRM_3 = 1 false = 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.
RM(x1, x2) = RM(x2)
cons(x1, x2) = cons(x2)
IFRM(x1, x2, x3) = IFRM(x1, 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

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

0 > false
0 > true
purge₁ > cons₁ > [RM₁, IFRM₂] > eq > false
purge₁ > cons₁ > [RM₁, IFRM₂] > eq > true
purge₁ > nil

Status:

RM₁: [1]
cons₁: [1]
IFRM₂: [1,2]
eq: multiset
false: multiset
true: multiset
0: multiset
purge₁: multiset
nil: multiset

(16) Obligation:

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

RM(cons(x)) → IFRM(eq, cons(x))
IFRM(false0, cons(x)) → RM(x)
IFRM(true0, cons(x)) → RM(x)

The TRS R consists of the following rules:

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

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

(17) MRRProof (EQUIVALENT transformation)

IFRM(false0, cons(x)) → RM(x)
IFRM(true0, cons(x)) → RM(x)

Strictly oriented rules of the TRS R:

ifrm(cons(x)) → rm(x)
purge0(cons(x)) → cons(purge0(rm(x)))

Used ordering: Polynomial interpretation [POLO]:

POL(IFRM(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(RM(x₁)) = 1 + x₁
POL(cons(x₁)) = 2 + x₁
POL(eq) = 0
POL(false0) = 0
POL(ifrm(x₁)) = x₁
POL(nil0) = 0
POL(purge0(x₁)) = 2·x₁
POL(rm(x₁)) = x₁
POL(true0) = 0

(18) Obligation:

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

RM(cons(x)) → IFRM(eq, cons(x))

The TRS R consists of the following rules:

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

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

(19) DependencyGraphProof (EQUIVALENT transformation)

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

(20) TRUE

(21) 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)
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
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)))
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(n, cons(m, x))) → PURGE(cons(m, cons(n, x)))

(22) 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))
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)))
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(EQ(x₁, x₂)) = 2 + 2·x₁·x₂
POL(IFRM(x₁, x₂, x₃)) = 2
POL(PURGE(x₁)) = 2 + 2·x₁
POL(RM(x₁, x₂)) = 2·x₁ + 2·x₁·x₂ + x₂
POL(cons(x₁, x₂)) = 1 + x₂
POL(eq(x₁, x₂)) = 0
POL(false) = 0
POL(ifrm(x₁, x₂, x₃)) = x₁ + x₃
POL(nil) = 0
POL(purge(x₁)) = 1 + 2·x₁
POL(rm(x₁, x₂)) = x₂
POL(s(x₁)) = 3 + 3·x₁
POL(true) = 0

(23) Obligation:

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

ifrm(true, n, cons(m, x)) → rm(n, x)
ifrm(false, n, cons(m, x)) → cons(m, rm(n, x))
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
rm(n, nil) → nil
rm(n, cons(m, x)) → ifrm(eq(n, m), n, cons(m, x))
purge(nil) → nil
purge(cons(n, cons(m, x))) → PURGE(cons(m, cons(n, x)))

(24) 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) DependencyGraphProof (EQUIVALENT transformation)

(20) TRUE

(21) Obligation:

(22) RelADPReductionPairProof (EQUIVALENT transformation)

(23) Obligation:

(24) DAbsisEmptyProof (EQUIVALENT transformation)

(25) YES