YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

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

1.1 Rule Removal

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

1.1.1 Dependency Pair Transformation

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

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals
[b#(x1)] =
0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[b(x1)] =
0 0 0 1
0 0 0 1
0 0 0 1
0 0 0 1
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[d(x1)] =
0 0 1 0
0 0 1 0
0 0 1 0
0 0 1 0
· x1 +
1 0 0 0
1 0 0 0
0 0 0 0
0 0 0 0
[d#(x1)] =
0 0 1 0
0 0 0 0
0 0 0 0
0 0 0 0
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[a(x1)] =
0 0 1 0
0 0 1 0
0 0 1 0
0 0 1 0
· x1 +
1 0 0 0
1 0 0 0
1 0 0 0
1 0 0 0
[a#(x1)] =
0 0 1 0
0 0 0 0
0 0 0 0
0 0 0 0
· x1 +
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
[c(x1)] =
0 0 1 0
0 1 0 0
0 0 0 1
1 0 0 0
· x1 +
1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
the pairs
a#(x0) → d#(x0)
d#(c(x0)) → b#(x0)
remain.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.