YES Termination proof of rt3-9.trs

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 (⇒, 42 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 4 ms)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔, 0 ms)
        ↳11 YES
    ↳12 RelADPP
        ↳13 RelADPCleverAfsProof (⇒, 24 ms)
        ↳14 QDP
        ↳15 MRRProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 PisEmptyProof (⇔, 0 ms)
        ↳18 YES
    ↳19 RelADPP
        ↳20 RelADPReductionPairProof (⇔, 27 ms)
        ↳21 RelADPP
        ↳22 RelADPDepGraphProof (⇔, 0 ms)
        ↳23 TRUE
    ↳24 RelADPP
        ↳25 RelADPReductionPairProof (⇔, 19 ms)
        ↳26 RelADPP
        ↳27 RelADPReductionPairProof (⇔, 0 ms)
        ↳28 RelADPP
        ↳29 RelADPReductionPairProof (⇔, 0 ms)
        ↳30 RelADPP
        ↳31 DAbsisEmptyProof (⇔, 0 ms)
        ↳32 YES
    ↳33 RelADPP
        ↳34 RelADPReductionPairProof (⇔, 26 ms)
        ↳35 RelADPP
        ↳36 RelADPDepGraphProof (⇔, 0 ms)
        ↳37 TRUE

(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:

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

and relative ADPs:

f(x, y) → f(x, r(y))
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 = f_2 = 0 m_1 = 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)  =  M(x1)
r(x1)  =  r(x1)
m(x1)  =  x1
f(x1, x2)  =  x2
l(x1)  =  l(x1)

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

trivial

Status:
M1: multiset
r1: multiset
l1: 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))
f(y) → f(m0(y))
f(y) → f(r0(y))
l0(m0(x)) → m0(l0(x))
b0 → l0(b0)

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

(8) 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)  =  x1
r0(x1)  =  r0(x1)
m0(x1)  =  x1
f(x1)  =  f
l0(x1)  =  l0
b0  =  b0

Recursive path order with status [RPO].
Quasi-Precedence:
f > r01
b0 > l0

Status:
r01: multiset
f: multiset
l0: 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))
f(y) → f(m0(y))
f(y) → f(r0(y))
l0(m0(x)) → m0(l0(x))
b0 → l0(b0)

(9) Obligation:

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

m0(r0(x)) → r0(m0(x))
f(y) → f(m0(y))
f(y) → f(r0(y))
l0(m0(x)) → m0(l0(x))
b0 → l0(b0)

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:

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

and relative ADPs:

f(x, y) → f(x, r(y))
b → l(b)

(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:L_1 = r_1 = 0 b = m_1 = f_2 = 1 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)  =  L(x1)
m(x1)  =  m(x1)
r(x1)  =  r
f(x1, x2)  =  f(x1)
l(x1)  =  x1

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

L1 > m1
r > m1
f1 > m1

Status:
L1: multiset
m1: multiset
r: multiset
f1: [1]

(14) 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(r) → r
f(m0(x)) → f(x)
f(x) → f(x)
l0(m0(x)) → m0(l0(x))
b0 → l0(b0)

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

(15) 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:

m0(r) → r
f(m0(x)) → f(x)
l0(m0(x)) → m0(l0(x))

Used ordering: Polynomial interpretation [POLO]:

POL(L0(x1)) = x1   
POL(b0) = 0   
POL(f(x1)) = x1   
POL(l0(x1)) = 2·x1   
POL(m0(x1)) = 2 + 2·x1   
POL(r) = 1   

(16) Obligation:

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

f(x) → f(x)
b0 → l0(b0)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) YES

(19) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

f(x, y) → F(x, r(y))
b → l(b)

(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:


f(m(x), y) → F(x, m(y))
f(m(x), y) → f(x, m(y))

Relative ADPs:

b → l(b)


The remaining rules can at least be oriented weakly:
Absolute ADPs:

m(r(x)) → r(m(x))
l(m(x)) → m(l(x))

Relative ADPs:

f(x, y) → F(x, r(y))


Ordered with Polynomial interpretation [POLO]:

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

(21) Obligation:

Relative ADP Problem with
absolute ADPs:

m(r(x)) → r(m(x))
l(m(x)) → m(l(x))

and relative ADPs:

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

(22) RelADPDepGraphProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

b → L(B)
f(x, y) → f(x, r(y))

(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:


f(m(x), y) → f(x, m(y))

Relative ADPs:

f(x, y) → f(x, r(y))


The remaining rules can at least be oriented weakly:
Absolute ADPs:

m(r(x)) → r(m(x))
l(m(x)) → m(L(x))
l(m(x)) → m(l(x))

Relative ADPs:

b → L(B)


Ordered with Polynomial interpretation [POLO]:

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

(26) Obligation:

Relative ADP Problem with
absolute ADPs:

m(r(x)) → r(m(x))
l(m(x)) → m(L(x))
l(m(x)) → m(l(x))

and relative ADPs:

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

(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:


m(r(x)) → r(m(x))

Relative ADPs:

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


The remaining rules can at least be oriented weakly:
Absolute ADPs:

l(m(x)) → m(L(x))
l(m(x)) → m(l(x))

Relative ADPs:

b → L(B)


Ordered with Polynomial interpretation [POLO]:

POL(B) = 0   
POL(F(x1, x2)) = 0   
POL(L(x1)) = 0   
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)) = 0   

(28) Obligation:

Relative ADP Problem with
absolute ADPs:

l(m(x)) → m(L(x))
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))

(29) 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))
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))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

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

(30) 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))

(31) DAbsisEmptyProof (EQUIVALENT transformation)

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

(32) YES

(33) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

b → L(B)
f(x, y) → f(x, r(y))

(34) 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))
l(m(x)) → m(l(x))

Relative ADPs:

f(x, y) → f(x, r(y))


The remaining rules can at least be oriented weakly:
Absolute ADPs:

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

Relative ADPs:

b → L(B)


Ordered with Polynomial interpretation [POLO]:

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

(35) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

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

(36) RelADPDepGraphProof (EQUIVALENT transformation)

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

(37) TRUE