YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

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

1.1 Dependency Pair Transformation

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

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the arctic semiring over the integers
[a#(x1)] = 8 · x1 + -∞
[c(x1)] = 0 · x1 + -∞
[b#(x1)] = 10 · x1 + 10
[c#(x1)] = 8 · x1 + -∞
[b(x1)] = 2 · x1 + 8
[a(x1)] = 0 · x1 + 0
together with the usable rules
b(c(c(x0))) → b(c(a(x0)))
a(b(c(a(x0)))) → c(c(b(x0)))
c(a(b(x0))) → a(c(b(a(x0))))
a(c(b(x0))) → b(a(c(x0)))
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(c(c(x0))) → b#(c(a(x0)))
a#(b(c(a(x0)))) → b#(x0)
a#(b(c(a(x0)))) → c#(b(x0))
a#(b(c(a(x0)))) → c#(c(b(x0)))
c#(a(b(x0))) → b#(a(x0))
c#(a(b(x0))) → c#(b(a(x0)))
c#(a(b(x0))) → a#(c(b(a(x0))))
a#(c(b(x0))) → b#(a(c(x0)))
remain.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.