YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

a(x0) → b(b(x0))
c(b(x0)) → d(x0)
e(b(x0)) → c(c(x0))
d(b(x0)) → b(f(x0))
f(x0) → a(e(x0))
c(x0) → x0
a(a(x0)) → f(x0)

Proof

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the arctic semiring over the integers
[e(x1)] =
0 -∞ 0
0 -∞ 0
0 -∞ 0
· x1 +
-∞ -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[b(x1)] =
0 0 0
0 0 1
0 -∞ 0
· x1 +
-∞ -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[d(x1)] =
0 0 1
0 0 1
0 0 1
· x1 +
-∞ -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[f(x1)] =
1 -∞ 1
1 0 1
0 -∞ 0
· x1 +
-∞ -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[a(x1)] =
1 1 1
1 0 1
0 0 0
· x1 +
-∞ -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[c(x1)] =
0 0 -∞
-∞ 0 -∞
0 0 0
· x1 +
-∞ -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
the rules
a(x0) → b(b(x0))
c(b(x0)) → d(x0)
e(b(x0)) → c(c(x0))
d(b(x0)) → b(f(x0))
f(x0) → a(e(x0))
c(x0) → x0
remain.

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(b(x0)) → d#(x0)
e#(b(x0)) → c#(x0)
e#(b(x0)) → c#(c(x0))
d#(b(x0)) → f#(x0)
f#(x0) → e#(x0)
f#(x0) → a#(e(x0))

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.