Termination proof of AG_#3.1

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 (⇒, 38 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 MRRProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 QDPOrderProof (⇔, 0 ms)

        ↳13 QDP

        ↳14 PisEmptyProof (⇔, 0 ms)

        ↳15 YES

    ↳16 RelADPP

        ↳17 RelADPCleverAfsProof (⇒, 58 ms)

        ↳18 QDP

        ↳19 MRRProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 QDPOrderProof (⇔, 0 ms)

        ↳22 QDP

        ↳23 PisEmptyProof (⇔, 0 ms)

        ↳24 YES

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))

The relative TRS consists of the following S rules:

rand(x) → rand(s(x))
rand(x) → 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:

minus(x, 0) → x
minus(s(x), s(y)) → MINUS(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))
quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

and relative ADPs:

rand(x) → RAND(s(x))
rand(x) → x

(3) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
2 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 2 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x
minus(s(x), s(y)) → MINUS(x, y)
quot(0, s(y)) → 0

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → 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 = MINUS_2 = 0 = minus_2 = 1 rand_1 = quot_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.
MINUS(x1, x2) = MINUS(x1, x2)
s(x1) = s(x1)
quot(x1, x2) = quot(x1, x2)
minus(x1, x2) = minus(x1)
0 = 0

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

MINUS₂ > minus₁
[quot₂, 0] > s₁ > minus₁

Status:

MINUS₂: [2,1]
s₁: [1]
quot₂: [1,2]
minus₁: multiset
0: multiset

(7) Obligation:

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

MINUS0(s0(x), s0(y)) → MINUS0(x, y)

The TRS R consists of the following rules:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
minus(x) → x
rand0(x) → rand0(s0(x))
rand0(x) → x
quot0(00, s0(y)) → 00

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:

rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(MINUS0(x₁, x₂)) = x₁ + x₂
POL(minus(x₁)) = x₁
POL(quot0(x₁, x₂)) = x₁ + x₂
POL(rand0(x₁)) = 2 + x₁
POL(s0(x₁)) = x₁

(9) Obligation:

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

MINUS0(s0(x), s0(y)) → MINUS0(x, y)

The TRS R consists of the following rules:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
minus(x) → x
rand0(x) → rand0(s0(x))
quot0(00, s0(y)) → 00

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

(10) MRRProof (EQUIVALENT transformation)

quot0(00, s0(y)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1
POL(MINUS0(x₁, x₂)) = x₁ + x₂
POL(minus(x₁)) = x₁
POL(quot0(x₁, x₂)) = 2·x₁ + 2·x₂
POL(rand0(x₁)) = x₁
POL(s0(x₁)) = x₁

(11) Obligation:

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

MINUS0(s0(x), s0(y)) → MINUS0(x, y)

The TRS R consists of the following rules:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
minus(x) → x
rand0(x) → rand0(s0(x))

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

MINUS0(s0(x), s0(y)) → MINUS0(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS0(x1, x2) = MINUS0(x2)
s0(x1) = s0(x1)
quot0(x1, x2) = quot0(x1, x2)
minus(x1) = x1
rand0(x1) = rand0

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

MINUS0₁ > s0₁
quot0₂ > s0₁
rand0 > s0₁

Status:

MINUS0₁: multiset
s0₁: [1]
quot0₂: [1,2]
rand0: []

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
minus(x) → x
rand0(x) → rand0(s0(x))

(13) Obligation:

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

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
minus(x) → x
rand0(x) → rand0(s0(x))

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) YES

(16) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
quot(0, s(y)) → 0

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(17) 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 = QUOT_2 = 0 = minus_2 = 1 rand_1 = quot_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.
QUOT(x1, x2) = QUOT(x1, x2)
s(x1) = s(x1)
minus(x1, x2) = x1
quot(x1, x2) = quot(x1, x2)
0 = 0

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

QUOT₂ > s₁
[quot₂, 0] > s₁

Status:

QUOT₂: multiset
s₁: multiset
quot₂: [2,1]
0: multiset

(18) Obligation:

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

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
minus(x) → x
rand0(x) → rand0(s0(x))
rand0(x) → x
quot0(00, s0(y)) → 00

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

(19) MRRProof (EQUIVALENT transformation)

rand0(x) → x
quot0(00, s0(y)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(QUOT0(x₁, x₂)) = x₁ + x₂
POL(minus(x₁)) = x₁
POL(quot0(x₁, x₂)) = 1 + x₁ + x₂
POL(rand0(x₁)) = 2 + x₁
POL(s0(x₁)) = x₁

(20) Obligation:

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

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
minus(x) → x
rand0(x) → rand0(s0(x))

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( quot0(x₁, x₂) ) = 2x₁

POL( s0(x₁) ) = 2x₁ + 1

POL( minus(x₁) ) = x₁

POL( rand0(x₁) ) = max{0, -2}

POL( QUOT0(x₁, x₂) ) = 2x₁ + 2x₂ + 2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
minus(x) → x
rand0(x) → rand0(s0(x))

(22) Obligation:

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

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
minus(x) → x
rand0(x) → rand0(s0(x))

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(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) QDPOrderProof (EQUIVALENT transformation)

(13) Obligation:

(14) PisEmptyProof (EQUIVALENT transformation)

(15) YES

(16) Obligation:

(17) RelADPCleverAfsProof (SOUND transformation)

(18) Obligation:

(19) MRRProof (EQUIVALENT transformation)

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) PisEmptyProof (EQUIVALENT transformation)

(24) YES