YES Termination proof of AG_#3.47_rand.trs

(0) Obligation:

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

f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))

The relative TRS consists of the following S rules:

rand(x) → rand(s(x))
rand(x) → x

(1) RelTRSRRRProof (EQUIVALENT transformation)

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

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

POL(c(x1)) =
/0\
\1/
+
/10\
\11/
·x1

POL(s(x1)) =
/0\
\0/
+
/10\
\00/
·x1

POL(rand(x1)) =
/1\
\1/
+
/10\
\01/
·x1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

f(x, c(y)) → f(x, s(f(y, y)))
Rules from S:

rand(x) → x


(2) Obligation:

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

f(s(x), y) → f(x, s(c(y)))

The relative TRS consists of the following S rules:

rand(x) → rand(s(x))

(3) RelTRSRRRProof (EQUIVALENT transformation)

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

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

POL(s(x1)) =
/0\
\1/
+
/10\
\01/
·x1

POL(c(x1)) =
/0\
\1/
+
/10\
\00/
·x1

POL(rand(x1)) =
/1\
\1/
+
/10\
\00/
·x1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

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


(4) Obligation:

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

rand(x) → rand(s(x))

(5) RIsEmptyProof (EQUIVALENT transformation)

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

(6) YES