Certification Problem
Input (TPDB TRS_Standard/Der95/02)
The rewrite relation of the following TRS is considered.
|
.(1,x) |
→ |
x |
(1) |
|
.(x,1) |
→ |
x |
(2) |
|
.(i(x),x) |
→ |
1 |
(3) |
|
.(x,i(x)) |
→ |
1 |
(4) |
|
i(1) |
→ |
1 |
(5) |
|
i(i(x)) |
→ |
x |
(6) |
|
.(i(y),.(y,z)) |
→ |
z |
(7) |
|
.(y,.(i(y),z)) |
→ |
z |
(8) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [.(x1, x2)] |
= |
2 + 1 · x1 + 2 · x2
|
| [1] |
= |
1 |
| [i(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
.(1,x) |
→ |
x |
(1) |
|
.(x,1) |
→ |
x |
(2) |
|
.(i(x),x) |
→ |
1 |
(3) |
|
.(x,i(x)) |
→ |
1 |
(4) |
|
.(i(y),.(y,z)) |
→ |
z |
(7) |
|
.(y,.(i(y),z)) |
→ |
z |
(8) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(1) |
= |
0 |
|
weight(1) |
= |
1 |
|
|
|
| prec(i) |
= |
1 |
|
weight(i) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
i(1) |
→ |
1 |
(5) |
|
i(i(x)) |
→ |
x |
(6) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.