YES Termination proof of rt2-5.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPReductionPairProof (⇔, 53 ms)
↳4 RelADPP
↳5 DAbsisEmptyProof (⇔, 0 ms)
↳6 YES

(0) Obligation:

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

f(f(x)) → x

The relative TRS consists of the following S rules:

f(x) → g(f(g(f(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(f(x)) → x

and relative ADPs:

f(x) → g(F(g(F(x))))

(3) 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(f(x)) → x


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(F(x1)) = x1   
POL(f(x1)) = 1 + x1   
POL(g(x1)) = 0   

(4) Obligation:

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

f(f(x)) → x
f(x) → g(F(g(F(x))))

(5) DAbsisEmptyProof (EQUIVALENT transformation)

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

(6) YES