YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

a(x0) → g(d(x0))
b(b(b(x0))) → c(d(c(x0)))
b(b(x0)) → a(g(g(x0)))
c(d(x0)) → g(g(x0))
g(g(g(x0))) → b(b(x0))

Proof

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[c(x1)] = 4 · x1 + -∞
[d(x1)] = 0 · x1 + -∞
[b(x1)] = 3 · x1 + -∞
[a(x1)] = 2 · x1 + -∞
[g(x1)] = 2 · x1 + -∞
the rules
a(x0) → g(d(x0))
b(b(x0)) → a(g(g(x0)))
c(d(x0)) → g(g(x0))
g(g(g(x0))) → b(b(x0))
remain.

1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[c(x1)] =
1 2
-∞ 3
· x1 +
-∞ -∞
-∞ -∞
[d(x1)] =
0 0
0 0
· x1 +
-∞ -∞
-∞ -∞
[b(x1)] =
0 0
0 0
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 0
0 0
· x1 +
-∞ -∞
-∞ -∞
[g(x1)] =
0 0
0 -∞
· x1 +
-∞ -∞
-∞ -∞
the rules
a(x0) → g(d(x0))
b(b(x0)) → a(g(g(x0)))
g(g(g(x0))) → b(b(x0))
remain.

1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[d(x1)] =
0 -∞
0 -∞
· x1 +
-∞ -∞
-∞ -∞
[b(x1)] =
0 1
1 3
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
2 0
3 1
· x1 +
-∞ -∞
-∞ -∞
[g(x1)] =
0 0
0 2
· x1 +
-∞ -∞
-∞ -∞
the rules
b(b(x0)) → a(g(g(x0)))
g(g(g(x0))) → b(b(x0))
remain.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[b(x1)] =
1 0
0 2
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 -∞
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
[g(x1)] =
0 0
2 2
· x1 +
-∞ -∞
-∞ -∞
the rule
b(b(x0)) → a(g(g(x0)))
remains.

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[b(x1)] =
2 3
3 0
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 -∞
3 3
· x1 +
-∞ -∞
-∞ -∞
[g(x1)] =
0 0
1 0
· x1 +
-∞ -∞
-∞ -∞
all rules could be removed.

1.1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.