YES Termination proof of AG_#3.56_rand.trs

(0) Obligation:

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

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

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:
Polynomial interpretation [POLO]:

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

f(x) → x
Rules from S:

rand(x) → x


(2) Obligation:

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

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

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(g(x1)) =
/1\
\0/
+
/10\
\00/
·x1

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

POL(s(x1)) =
/0\
\0/
+
/10\
\00/
·x1

POL(f(x1)) =
/1\
\1/
+
/11\
\00/
·x1

POL(d(x1)) =
/0\
\0/
+
/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(f(x)) → f(d(f(x)))
Rules from S:
none


(4) Obligation:

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

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

The relative TRS consists of the following S rules:

rand(x) → rand(s(x))

(5) RelTRSRRRProof (EQUIVALENT transformation)

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

POL(g(x1)) =
/1\
\0/
+
/11\
\00/
·x1

POL(c(x1, x2)) =
/1\
\1/
+
/11\
\00/
·x1 +
/11\
\01/
·x2

POL(s(x1)) =
/0\
\1/
+
/10\
\01/
·x1

POL(f(x1)) =
/0\
\0/
+
/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:

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


(6) Obligation:

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

f(c(s(x), y)) → f(c(x, s(y)))

The relative TRS consists of the following S rules:

rand(x) → rand(s(x))

(7) RelTRSRRRProof (EQUIVALENT transformation)

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

POL(f(x1)) =
/0\
\0/
+
/10\
\00/
·x1

POL(c(x1, x2)) =
/0\
\1/
+
/11\
\01/
·x1 +
/10\
\00/
·x2

POL(s(x1)) =
/0\
\1/
+
/10\
\01/
·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(c(s(x), y)) → f(c(x, s(y)))
Rules from S:
none


(8) Obligation:

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

rand(x) → rand(s(x))

(9) RIsEmptyProof (EQUIVALENT transformation)

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

(10) YES