Termination proof of AProVE_24

Termination of the given RelTRS could be proven:

0 RelTRS

↳1 RelTRS S Cleaner (⇔, 0 ms)

↳2 RelTRS

↳3 RelTRStoRelADPProof (⇔, 0 ms)

↳4 RelADPP

↳5 RelADPDepGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 RelADPP

        ↳8 RelADPCleverAfsProof (⇒, 51 ms)

        ↳9 QDP

        ↳10 MRRProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 PisEmptyProof (⇔, 0 ms)

        ↳13 YES

    ↳14 RelADPP

        ↳15 RelADPCleverAfsProof (⇒, 39 ms)

        ↳16 QDP

        ↳17 MRRProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 PisEmptyProof (⇔, 0 ms)

        ↳20 YES

    ↳21 RelADPP

        ↳22 RelADPReductionPairProof (⇔, 50 ms)

        ↳23 RelADPP

        ↳24 DAbsisEmptyProof (⇔, 0 ms)

        ↳25 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)
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
half(s(s(x))) → div(x, s(s(0)))

The relative TRS consists of the following S rules:

gen(s(x), b) → gen(s(x), b)
gen(s(x), b) → gen(x, a)
gen(0, a) → gen(0, a)
gen(s(x), a) → c(s(s(0)), gen(s(x), b))
gen(s(x), a) → c(half(s(s(0))), gen(x, a))

(1) RelTRS S Cleaner (EQUIVALENT transformation)

We have deleted all rules from S that have the shape t → t:

gen(s(x), b) → gen(s(x), b)
gen(0, a) → gen(0, a)

(2) 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)
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
half(s(s(x))) → div(x, s(s(0)))

The relative TRS consists of the following S rules:

gen(s(x), b) → gen(x, a)
gen(s(x), a) → c(s(s(0)), gen(s(x), b))
gen(s(x), a) → c(half(s(s(0))), gen(x, a))

(3) RelTRStoRelADPProof (EQUIVALENT transformation)

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

(4) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(x, 0) → x
minus(s(x), s(y)) → MINUS(x, y)
div(0, s(y)) → 0
div(s(x), s(y)) → s(DIV(minus(x, y), s(y)))
div(s(x), s(y)) → s(div(MINUS(x, y), s(y)))
half(s(s(x))) → DIV(x, s(s(0)))

and relative ADPs:

gen(s(x), b) → GEN(x, a)
gen(s(x), a) → c(s(s(0)), GEN(s(x), b))
gen(s(x), a) → c(HALF(s(s(0))), GEN(x, a))

(5) 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,
1 Lasso,
Result: This relative DT problem is equivalent to 3 subproblems.

(6) Complex Obligation (AND)

(7) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(s(x), s(y)) → MINUS(x, y)

and relative ADPs:

div(0, s(y)) → 0
gen(s(x), b) → gen(x, a)
half(s(s(x))) → div(x, s(s(0)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
gen(s(x), a) → c(s(s(0)), gen(s(x), b))
minus(x, 0) → x
gen(s(x), a) → c(half(s(s(0))), gen(x, a))

(8) 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 a = b = c_2 = half_1 = div_2 = 0 = minus_2 = 1 gen_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(x2)
s(x1) = s(x1)
div(x1, x2) = div(x1, x2)
minus(x1, x2) = x1
0 = 0

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

MINUS₁ > s₁
div₂ > s₁
0 > s₁

Status:

MINUS₁: [1]
s₁: multiset
div₂: [2,1]
0: multiset

(9) Obligation:

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

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

div0(00, s0(y)) → 00
gen0(s0(x), b0) → gen0(x, a0)
half0(s0(s0(x))) → div0(x, s0(s0(00)))
div0(s0(x), s0(y)) → s0(div0(minus(x), s0(y)))
gen0(s0(x), a0) → c0(s0(s0(00)), gen0(s0(x), b0))
minus(s0(x)) → minus(x)
minus(x) → x
gen0(s0(x), a0) → c0(half0(s0(s0(00))), gen0(x, a0))

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

(10) 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:

MINUS(s0(y)) → MINUS(y)

Strictly oriented rules of the TRS R:

div0(00, s0(y)) → 00
minus(s0(x)) → minus(x)

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(MINUS(x₁)) = x₁
POL(a0) = 1
POL(b0) = 0
POL(c0(x₁, x₂)) = x₁ + x₂
POL(div0(x₁, x₂)) = x₁ + x₂
POL(gen0(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(half0(x₁)) = x₁
POL(minus(x₁)) = x₁
POL(s0(x₁)) = 1 + x₁

(11) Obligation:

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

gen0(s0(x), b0) → gen0(x, a0)
half0(s0(s0(x))) → div0(x, s0(s0(00)))
div0(s0(x), s0(y)) → s0(div0(minus(x), s0(y)))
gen0(s0(x), a0) → c0(s0(s0(00)), gen0(s0(x), b0))
minus(x) → x
gen0(s0(x), a0) → c0(half0(s0(s0(00))), gen0(x, a0))

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:

div(s(x), s(y)) → s(DIV(minus(x, y), s(y)))

and relative ADPs:

div(0, s(y)) → 0
gen(s(x), b) → gen(x, a)
half(s(s(x))) → div(x, s(s(0)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
gen(s(x), a) → c(s(s(0)), gen(s(x), b))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
gen(s(x), a) → c(half(s(s(0))), gen(x, a))

(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 = a = b = c_2 = half_1 = div_2 = 1 0 = minus_2 = 1 gen_2 = DIV_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.
DIV(x1, x2) = DIV(x1, x2)
s(x1) = s(x1)
minus(x1, x2) = x1
0 = 0
div(x1, x2) = div(x1)

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

0 > [DIV₂, s₁, div₁]

Status:

DIV₂: multiset
s₁: multiset
0: multiset
div₁: multiset

(16) Obligation:

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

DIV0(s0(x), s0(y)) → DIV0(minus(x), s0(y))

The TRS R consists of the following rules:

div(00) → 00
gen0(s0(x), b0) → gen0(x, a0)
half0(s0(s0(x))) → div(x)
div(s0(x)) → s0(div(minus(x)))
gen0(s0(x), a0) → c0(s0(s0(00)), gen0(s0(x), b0))
minus(s0(x)) → minus(x)
minus(x) → x
gen0(s0(x), a0) → c0(half0(s0(s0(00))), gen0(x, a0))

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

(17) MRRProof (EQUIVALENT transformation)

DIV0(s0(x), s0(y)) → DIV0(minus(x), s0(y))

Strictly oriented rules of the TRS R:

half0(s0(s0(x))) → div(x)
minus(s0(x)) → minus(x)

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(DIV0(x₁, x₂)) = x₁ + x₂
POL(a0) = 1
POL(b0) = 0
POL(c0(x₁, x₂)) = x₁ + x₂
POL(div(x₁)) = x₁
POL(gen0(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(half0(x₁)) = x₁
POL(minus(x₁)) = x₁
POL(s0(x₁)) = 1 + x₁

(18) Obligation:

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

div(00) → 00
gen0(s0(x), b0) → gen0(x, a0)
div(s0(x)) → s0(div(minus(x)))
gen0(s0(x), a0) → c0(s0(s0(00)), gen0(s0(x), b0))
minus(x) → x
gen0(s0(x), a0) → c0(half0(s0(s0(00))), gen0(x, a0))

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) YES

(21) Obligation:

Relative ADP Problem with
absolute ADPs:

half(s(s(x))) → DIV(x, s(s(0)))

and relative ADPs:

div(0, s(y)) → 0
gen(s(x), a) → c(s(s(0)), GEN(s(x), b))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
gen(s(x), b) → GEN(x, a)
gen(s(x), a) → c(HALF(s(s(0))), GEN(x, a))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → 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:

half(s(s(x))) → DIV(x, s(s(0)))

Relative ADPs:

div(0, s(y)) → 0
gen(s(x), b) → GEN(x, a)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0
POL(DIV(x₁, x₂)) = 0
POL(GEN(x₁, x₂)) = 2·x₁ + 3·x₂
POL(HALF(x₁)) = x₁
POL(MINUS(x₁, x₂)) = 2 + 2·x₁·x₂
POL(a) = 2
POL(b) = 2
POL(c(x₁, x₂)) = 0
POL(div(x₁, x₂)) = 2·x₁
POL(gen(x₁, x₂)) = 3·x₂
POL(half(x₁)) = 2 + 3·x₁
POL(minus(x₁, x₂)) = x₁
POL(s(x₁)) = 3 + x₁

(23) Obligation:

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

div(0, s(y)) → 0
gen(s(x), a) → c(s(s(0)), GEN(s(x), b))
gen(s(x), b) → gen(x, a)
half(s(s(x))) → div(x, s(s(0)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
gen(s(x), a) → c(HALF(s(s(0))), GEN(x, a))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

(24) DAbsisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) RelTRS S Cleaner (EQUIVALENT transformation)

(2) Obligation:

(3) RelTRStoRelADPProof (EQUIVALENT transformation)

(4) Obligation:

(5) RelADPDepGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) RelADPCleverAfsProof (SOUND 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) PisEmptyProof (EQUIVALENT transformation)

(20) YES

(21) Obligation:

(22) RelADPReductionPairProof (EQUIVALENT transformation)

(23) Obligation:

(24) DAbsisEmptyProof (EQUIVALENT transformation)

(25) YES