YES Termination proof of rt3-7.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPReductionPairProof (⇔, 50 ms)
↳4 RelADPP
↳5 RelADPReductionPairProof (⇔, 0 ms)
↳6 RelADPP
↳7 DAbsisEmptyProof (⇔, 0 ms)
↳8 YES

(0) Obligation:

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

f(b(x), y) → f(x, b(y))
f(x, a(y)) → f(a(x), y)

The relative TRS consists of the following S rules:

f(x, y) → f(x, b(y))
f(x, y) → f(a(x), y)

(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(b(x), y) → F(x, b(y))
f(x, a(y)) → F(a(x), y)

and relative ADPs:

f(x, y) → F(x, b(y))
f(x, y) → F(a(x), y)

(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(x, a(y)) → F(a(x), y)


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

f(b(x), y) → F(x, b(y))

Relative ADPs:

f(x, y) → F(a(x), y)
f(x, y) → F(x, b(y))


Ordered with Polynomial interpretation [POLO]:

POL(F(x1, x2)) = 3·x2   
POL(a(x1)) = 3 + x1   
POL(b(x1)) = x1   
POL(f(x1, x2)) = 0   

(4) Obligation:

Relative ADP Problem with
absolute ADPs:

f(b(x), y) → F(x, b(y))

and relative ADPs:

f(x, y) → F(a(x), y)
f(x, y) → F(x, b(y))
f(x, a(y)) → f(a(x), y)

(5) 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(b(x), y) → F(x, b(y))

Relative ADPs:

f(x, a(y)) → f(a(x), y)


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(F(x1, x2)) = 2·x1   
POL(a(x1)) = 0   
POL(b(x1)) = 2 + 2·x1   
POL(f(x1, x2)) = 0   

(6) Obligation:

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

f(x, y) → F(a(x), y)
f(x, y) → F(x, b(y))
f(b(x), y) → f(x, b(y))
f(x, a(y)) → f(a(x), y)

(7) DAbsisEmptyProof (EQUIVALENT transformation)

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

(8) YES