YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

f(x0) → n(c(n(a(x0))))
c(f(x0)) → f(n(a(c(x0))))
n(a(x0)) → c(x0)
c(c(x0)) → c(x0)
n(s(x0)) → f(s(s(x0)))
n(f(x0)) → f(n(x0))

Proof

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
f(x0) → a(n(c(n(x0))))
f(c(x0)) → c(a(n(f(x0))))
a(n(x0)) → c(x0)
c(c(x0)) → c(x0)
s(n(x0)) → s(s(f(x0)))
f(n(x0)) → n(f(x0))

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[s(x1)] =
1 0 1
1 0 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
0 1 0
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
[n(x1)] =
1 0 0
0 1 0
0 1 0
· x1 +
0 0 0
1 0 0
1 0 0
the rules
f(c(x0)) → c(a(n(f(x0))))
a(n(x0)) → c(x0)
c(c(x0)) → c(x0)
s(n(x0)) → s(s(f(x0)))
f(n(x0)) → n(f(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
[s(x1)] =
0 1
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 0
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
[c(x1)] =
0 2
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
[f(x1)] =
0 0
-∞ 0
· x1 +
-∞ -∞
-∞ -∞
[n(x1)] =
0 2
0 0
· x1 +
-∞ -∞
-∞ -∞
the rules
f(c(x0)) → c(a(n(f(x0))))
a(n(x0)) → c(x0)
c(c(x0)) → c(x0)
f(n(x0)) → n(f(x0))
remain.

1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
c(f(x0)) → f(n(a(c(x0))))
n(a(x0)) → c(x0)
c(c(x0)) → c(x0)
n(f(x0)) → f(n(x0))

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0
0 0
· x1 +
0 0
0 0
[c(x1)] =
1 0
1 0
· x1 +
0 0
0 0
[f(x1)] =
2 0
0 2
· x1 +
1 0
1 0
[n(x1)] =
1 2
1 0
· x1 +
0 0
0 0
the rules
c(f(x0)) → f(n(a(c(x0))))
n(a(x0)) → c(x0)
c(c(x0)) → c(x0)
remain.

1.1.1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
f(c(x0)) → c(a(n(f(x0))))
a(n(x0)) → c(x0)
c(c(x0)) → c(x0)

1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
1 0 0
1 0 0
[c(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
1 0 0
1 0 0
[f(x1)] =
1 0 1
0 1 1
1 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[n(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
the rules
f(c(x0)) → c(a(n(f(x0))))
c(c(x0)) → c(x0)
remain.

1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[a(x1)] = 0 · x1 + -∞
[c(x1)] = 8 · x1 + -∞
[f(x1)] = 2 · x1 + -∞
[n(x1)] = 0 · x1 + -∞
the rule
f(c(x0)) → c(a(n(f(x0))))
remains.

1.1.1.1.1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1.1.1.1.1 R is empty

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