YES Termination proof of Relative_05_rt3-9.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 (⇒, 44 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 6 ms)
        ↳11 QDP
        ↳12 PisEmptyProof (⇔, 0 ms)
        ↳13 YES
    ↳14 RelADPP
        ↳15 RelADPCleverAfsProof (⇒, 28 ms)
        ↳16 QDP
        ↳17 MRRProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 PisEmptyProof (⇔, 0 ms)
        ↳20 YES
    ↳21 RelADPP
        ↳22 RelADPReductionPairProof (⇔, 41 ms)
        ↳23 RelADPP
        ↳24 DAbsisEmptyProof (⇔, 0 ms)
        ↳25 YES
    ↳26 RelADPP
        ↳27 RelADPReductionPairProof (⇔, 33 ms)
        ↳28 RelADPP
        ↳29 DAbsisEmptyProof (⇔, 0 ms)
        ↳30 YES
    ↳31 RelADPP
        ↳32 RelADPReductionPairProof (⇔, 36 ms)
        ↳33 RelADPP
        ↳34 DAbsisEmptyProof (⇔, 0 ms)
        ↳35 YES

(0) Obligation:

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

l(m(x)) → m(l(x))
m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))

The relative TRS consists of the following S rules:

f(x, y) → f(x, r(y))
b → l(b)

(1) RelTRStoRelADPProof (EQUIVALENT transformation)

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

(2) Obligation:

Relative ADP Problem with
absolute ADPs:

l(m(x)) → M(l(x))
l(m(x)) → m(L(x))
m(r(x)) → r(M(x))
f(m(x), y) → F(x, m(y))
f(m(x), y) → f(x, M(y))

and relative ADPs:

f(x, y) → F(x, r(y))
b → L(B)

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

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

m(r(x)) → r(M(x))

and relative ADPs:

f(m(x), y) → f(x, m(y))
f(x, y) → f(x, r(y))
l(m(x)) → m(l(x))
b → l(b)

(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:r_1 = M_1 = b = m_1 = f_2 = l_1 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
M(x1)  =  x1
r(x1)  =  r(x1)

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

trivial

Status:
r1: multiset

(7) Obligation:

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

M0(r0(x)) → M0(x)

The TRS R consists of the following rules:

m0(r0(x)) → r0(m0(x))
f0(m0(x), y) → f0(x, m0(y))
f0(x, y) → f0(x, r0(y))
l0(m0(x)) → m0(l0(x))
b0 → l0(b0)

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:

f0(m0(x), y) → f0(x, m0(y))
l0(m0(x)) → m0(l0(x))

Used ordering: Polynomial interpretation [POLO]:

POL(M0(x1)) = 2·x1   
POL(b0) = 0   
POL(f0(x1, x2)) = 2·x1 + x2   
POL(l0(x1)) = 2·x1   
POL(m0(x1)) = 1 + x1   
POL(r0(x1)) = x1   

(9) Obligation:

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

M0(r0(x)) → M0(x)

The TRS R consists of the following rules:

m0(r0(x)) → r0(m0(x))
f0(x, y) → f0(x, r0(y))
b0 → l0(b0)

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

(10) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


M0(r0(x)) → M0(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
M0(x1)  =  M0(x1)
r0(x1)  =  r0(x1)
m0(x1)  =  m0(x1)
f0(x1, x2)  =  f0(x1)
b0  =  b0
l0(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[M01, r01, m01]

Status:
M01: [1]
r01: [1]
m01: [1]
f01: multiset
b0: multiset


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

m0(r0(x)) → r0(m0(x))
f0(x, y) → f0(x, r0(y))
b0 → l0(b0)

(11) Obligation:

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

m0(r0(x)) → r0(m0(x))
f0(x, y) → f0(x, r0(y))
b0 → l0(b0)

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:

l(m(x)) → m(L(x))

and relative ADPs:

m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))
f(x, y) → f(x, r(y))
l(m(x)) → m(l(x))
b → l(b)

(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:L_1 = r_1 = b = m_1 = f_2 = l_1 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
L(x1)  =  x1
m(x1)  =  m(x1)

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

trivial

Status:
m1: multiset

(16) Obligation:

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

L0(m0(x)) → L0(x)

The TRS R consists of the following rules:

m0(r0(x)) → r0(m0(x))
f0(m0(x), y) → f0(x, m0(y))
f0(x, y) → f0(x, r0(y))
l0(m0(x)) → m0(l0(x))
b0 → l0(b0)

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

(17) 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:

L0(m0(x)) → L0(x)

Strictly oriented rules of the TRS R:

f0(m0(x), y) → f0(x, m0(y))
l0(m0(x)) → m0(l0(x))

Used ordering: Polynomial interpretation [POLO]:

POL(L0(x1)) = x1   
POL(b0) = 0   
POL(f0(x1, x2)) = 2·x1 + x2   
POL(l0(x1)) = 2·x1   
POL(m0(x1)) = 2 + x1   
POL(r0(x1)) = x1   

(18) Obligation:

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

m0(r0(x)) → r0(m0(x))
f0(x, y) → f0(x, r0(y))
b0 → l0(b0)

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:

f(m(x), y) → F(x, m(y))

and relative ADPs:

f(x, y) → F(x, r(y))
m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))
l(m(x)) → m(l(x))
b → l(b)

(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:


f(m(x), y) → F(x, m(y))

Relative ADPs:

m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))
l(m(x)) → m(l(x))
b → l(b)


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(B) = 3   
POL(F(x1, x2)) = 2 + 3·x1   
POL(L(x1)) = 2 + 3·x12   
POL(M(x1)) = 3   
POL(b) = 0   
POL(f(x1, x2)) = 2·x1   
POL(l(x1)) = x1   
POL(m(x1)) = 1 + 2·x1   
POL(r(x1)) = 1   

(23) Obligation:

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

f(x, y) → F(x, r(y))
f(m(x), y) → f(x, m(y))
m(r(x)) → r(m(x))
l(m(x)) → m(l(x))
b → l(b)

(24) DAbsisEmptyProof (EQUIVALENT transformation)

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

(25) YES

(26) Obligation:

Relative ADP Problem with
absolute ADPs:

l(m(x)) → m(L(x))

and relative ADPs:

b → L(B)
m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))
f(x, y) → f(x, r(y))
l(m(x)) → m(l(x))

(27) 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:


l(m(x)) → m(L(x))

Relative ADPs:

m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))
f(x, y) → f(x, r(y))
l(m(x)) → m(l(x))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(B) = 0   
POL(F(x1, x2)) = 2   
POL(L(x1)) = 3·x1   
POL(M(x1)) = 2   
POL(b) = 0   
POL(f(x1, x2)) = 2·x1   
POL(l(x1)) = x1   
POL(m(x1)) = 3 + 2·x1   
POL(r(x1)) = 0   

(28) Obligation:

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

b → L(B)
m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))
f(x, y) → f(x, r(y))
l(m(x)) → m(l(x))

(29) DAbsisEmptyProof (EQUIVALENT transformation)

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

(30) YES

(31) Obligation:

Relative ADP Problem with
absolute ADPs:

l(m(x)) → M(l(x))

and relative ADPs:

b → L(B)
m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))
f(x, y) → f(x, r(y))
l(m(x)) → m(l(x))

(32) 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:


l(m(x)) → M(l(x))

Relative ADPs:

m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))
f(x, y) → f(x, r(y))
l(m(x)) → m(l(x))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(B) = 0   
POL(F(x1, x2)) = 2   
POL(L(x1)) = 3·x1   
POL(M(x1)) = 1   
POL(b) = 0   
POL(f(x1, x2)) = 3·x1   
POL(l(x1)) = 2·x1   
POL(m(x1)) = 2 + 2·x1   
POL(r(x1)) = 1   

(33) Obligation:

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

b → L(B)
m(r(x)) → r(m(x))
f(m(x), y) → f(x, m(y))
f(x, y) → f(x, r(y))
l(m(x)) → m(l(x))

(34) DAbsisEmptyProof (EQUIVALENT transformation)

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

(35) YES