YES Termination proof of rt2-7.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPDepGraphProof (⇔, 0 ms)
↳4 RelADPP
↳5 RelADPCleverAfsProof (⇒, 58 ms)
↳6 QDP
↳7 MRRProof (⇔, 0 ms)
↳8 QDP
↳9 QDPOrderProof (⇔, 0 ms)
↳10 QDP
↳11 PisEmptyProof (⇔, 0 ms)
↳12 YES

(0) Obligation:

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

f(g(f(x))) → f(g(g(g(f(x)))))

The relative TRS consists of the following S rules:

g(x) → g(g(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:

f(g(f(x))) → F(g(g(g(f(x)))))
f(g(f(x))) → f(G(g(g(f(x)))))
f(g(f(x))) → f(g(G(g(f(x)))))
f(g(f(x))) → f(g(g(G(f(x)))))
f(g(f(x))) → f(g(g(g(F(x)))))

and relative ADPs:

g(x) → G(G(x))

(3) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
1 SCC with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 1 subproblem.

(4) Obligation:

Relative ADP Problem with
absolute ADPs:

f(g(f(x))) → f(g(g(g(f(x)))))
f(g(f(x))) → f(g(g(g(F(x)))))
f(g(f(x))) → F(g(g(g(f(x)))))

and relative ADPs:

g(x) → g(g(x))

(5) 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:f_1 = F_1 = g_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.
F(x1)  =  F(x1)
g(x1)  =  x1
f(x1)  =  f(x1)

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

F1 > f1

Status:
F1: [1]
f1: multiset

(6) Obligation:

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

F0(g0(f0(x))) → F0(x)
F0(g0(f0(x))) → F0(g0(g0(g0(f0(x)))))

The TRS R consists of the following rules:

f0(g0(f0(x))) → f0(g0(g0(g0(f0(x)))))
g0(x) → g0(g0(x))

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

(7) 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:

F0(g0(f0(x))) → F0(x)


Used ordering: Polynomial interpretation [POLO]:

POL(F0(x1)) = 2·x1   
POL(f0(x1)) = 2 + 2·x1   
POL(g0(x1)) = x1   

(8) Obligation:

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

F0(g0(f0(x))) → F0(g0(g0(g0(f0(x)))))

The TRS R consists of the following rules:

f0(g0(f0(x))) → f0(g0(g0(g0(f0(x)))))
g0(x) → g0(g0(x))

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

(9) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


F0(g0(f0(x))) → F0(g0(g0(g0(f0(x)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(F0(x1)) = x1   
POL(f0(x1)) = [1/4]   
POL(g0(x1)) = [1/4]x1   
The value of delta used in the strict ordering is 15/256.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f0(g0(f0(x))) → f0(g0(g0(g0(f0(x)))))
g0(x) → g0(g0(x))

(10) Obligation:

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

f0(g0(f0(x))) → f0(g0(g0(g0(f0(x)))))
g0(x) → g0(g0(x))

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) YES