YES Termination proof of rt3-7.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRSRRRProof (⇔, 43 ms)
↳2 RelTRS
↳3 RelTRSRRRProof (⇔, 0 ms)
↳4 RelTRS
↳5 RIsEmptyProof (⇔, 0 ms)
↳6 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) RelTRSRRRProof (EQUIVALENT transformation)

We used the following monotonic ordering for rule removal:
Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(f(x1, x2)) =
/1\
\0/
 +
/10\
\00/
·x1 +
/11\
\11/
·x2

POL(b(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

POL(a(x1)) =
/0\
\1/
 +
/10\
\11/
·x1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

f(x, a(y)) → f(a(x), y)
Rules from S:
none


(2) Obligation:

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

f(b(x), y) → f(x, b(y))

The relative TRS consists of the following S rules:

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

(3) RelTRSRRRProof (EQUIVALENT transformation)

We used the following monotonic ordering for rule removal:
Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(f(x1, x2)) =
/1\
\0/
 +
/11\
\11/
·x1 +
/10\
\00/
·x2

POL(b(x1)) =
/0\
\1/
 +
/10\
\11/
·x1

POL(a(x1)) =
/0\
\0/
 +
/10\
\00/
·x1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

f(b(x), y) → f(x, b(y))
Rules from S:
none


(4) Obligation:

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

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

(5) RIsEmptyProof (EQUIVALENT transformation)

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

(6) YES