YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
0(0(0(0(x0)))) |
→ |
0(1(1(1(x0)))) |
1(1(0(1(x0)))) |
→ |
0(0(0(0(x0)))) |
Proof
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
0#(0(0(0(x0)))) |
→ |
1#(x0) |
0#(0(0(0(x0)))) |
→ |
1#(1(x0)) |
0#(0(0(0(x0)))) |
→ |
1#(1(1(x0))) |
0#(0(0(0(x0)))) |
→ |
0#(1(1(1(x0)))) |
1#(1(0(1(x0)))) |
→ |
0#(x0) |
1#(1(0(1(x0)))) |
→ |
0#(0(x0)) |
1#(1(0(1(x0)))) |
→ |
0#(0(0(x0))) |
1#(1(0(1(x0)))) |
→ |
0#(0(0(0(x0)))) |
1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the arctic semiring over the integers
[1(x1)] |
= |
1 ·
x1 +
-∞
|
[1#(x1)] |
= |
0 ·
x1 +
-∞
|
[0(x1)] |
= |
1 ·
x1 +
-∞
|
[0#(x1)] |
= |
0 ·
x1 +
-∞
|
together with the usable
rules
0(0(0(0(x0)))) |
→ |
0(1(1(1(x0)))) |
1(1(0(1(x0)))) |
→ |
0(0(0(0(x0)))) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
0#(0(0(0(x0)))) |
→ |
0#(1(1(1(x0)))) |
1#(1(0(1(x0)))) |
→ |
0#(0(0(0(x0)))) |
remain.
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.