YES Termination proof of rt2-7.trs

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRS Reverse (⇔, 0 ms)
↳2 RelTRS
↳3 RootLabelingProof (⇔, 0 ms)
↳4 RelTRS
↳5 RelTRSRRRProof (⇔, 112 ms)
↳6 RelTRS
↳7 RIsEmptyProof (⇔, 0 ms)
↳8 YES

(0) Obligation:

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

f(g(f(x))) → f(g(g(g(f(x)))))

The relative TRS consists of the following S rules:

g(x) → g(g(x))

(1) RelTRS Reverse (EQUIVALENT transformation)

We have reversed the following relative TRS [REVERSE]:
The set of rules R is

f(g(f(x))) → f(g(g(g(f(x)))))

The set of rules S is

g(x) → g(g(x))

We have obtained the following relative TRS:
The set of rules R is

f(g(f(x))) → f(g(g(g(f(x)))))

The set of rules S is

g(x) → g(g(x))

(2) Obligation:

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

f(g(f(x))) → f(g(g(g(f(x)))))

The relative TRS consists of the following S rules:

g(x) → g(g(x))

(3) RootLabelingProof (EQUIVALENT transformation)

We used plain root labeling [ROOTLAB] with the following heuristic: LabelAll: All function symbols get labeled

(4) Obligation:

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

f_{g_1}(g_{f_1}(f_{f_1}(x))) → f_{g_1}(g_{g_1}(g_{g_1}(g_{f_1}(f_{f_1}(x)))))
f_{g_1}(g_{f_1}(f_{g_1}(x))) → f_{g_1}(g_{g_1}(g_{g_1}(g_{f_1}(f_{g_1}(x)))))

The relative TRS consists of the following S rules:

g_{f_1}(x) → g_{g_1}(g_{f_1}(x))
g_{g_1}(x) → g_{g_1}(g_{g_1}(x))

(5) RelTRSRRRProof (EQUIVALENT transformation)

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

POL(f_{g_1}(x1)) =
/2\
\1/
 +
/22\
\22/
·x1

POL(g_{f_1}(x1)) =
/1\
\1/
 +
/22\
\00/
·x1

POL(f_{f_1}(x1)) =
/0\
\0/
 +
/10\
\10/
·x1

POL(g_{g_1}(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_{g_1}(g_{f_1}(f_{f_1}(x))) → f_{g_1}(g_{g_1}(g_{g_1}(g_{f_1}(f_{f_1}(x)))))
f_{g_1}(g_{f_1}(f_{g_1}(x))) → f_{g_1}(g_{g_1}(g_{g_1}(g_{f_1}(f_{g_1}(x)))))
Rules from S:
none


(6) Obligation:

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

g_{f_1}(x) → g_{g_1}(g_{f_1}(x))
g_{g_1}(x) → g_{g_1}(g_{g_1}(x))

(7) RIsEmptyProof (EQUIVALENT transformation)

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

(8) YES