YES Termination proof of rt1-5.trs

(0) Obligation:

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

s(a(x)) → s(b(x))
b(b(x)) → a(x)

The relative TRS consists of the following S rules:

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

(1) RelTRSRRRProof (EQUIVALENT transformation)

We used the following monotonic ordering for rule removal:
Polynomial interpretation [POLO]:

POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(f(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

b(b(x)) → a(x)
Rules from S:
none


(2) Obligation:

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

s(a(x)) → s(b(x))

The relative TRS consists of the following S rules:

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

(3) RelTRSRRRProof (EQUIVALENT transformation)

We used the following monotonic ordering for rule removal:
Polynomial interpretation [POLO]:

POL(a(x1)) = 1 + x1   
POL(b(x1)) = x1   
POL(f(x1, x2)) = x1 + x2   
POL(s(x1)) = x1   
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

s(a(x)) → s(b(x))
Rules from S:
none


(4) Obligation:

Relative term rewrite system:
R is empty.
The relative TRS consists of the following S rules:

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

(5) RIsEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(6) YES