Certification Problem
Input (TPDB TRS_Standard/Various_04/23)
The rewrite relation of the following TRS is considered.
|
g(0,f(x,x)) |
→ |
x |
(1) |
|
g(x,s(y)) |
→ |
g(f(x,y),0) |
(2) |
|
g(s(x),y) |
→ |
g(f(x,y),0) |
(3) |
|
g(f(x,y),0) |
→ |
f(g(x,0),g(y,0)) |
(4) |
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
| [g(x1, x2)] |
= |
2 · x1 + 2 · x2
|
| [0] |
= |
0 |
| [f(x1, x2)] |
= |
1 + 1 · x1 + 1 · x2
|
| [s(x1)] |
= |
1 + 1 · x1
|
all of the following rules can be deleted.
|
g(0,f(x,x)) |
→ |
x |
(1) |
|
g(f(x,y),0) |
→ |
f(g(x,0),g(y,0)) |
(4) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(0) |
= |
3 |
|
weight(0) |
= |
2 |
|
|
|
| prec(s) |
= |
1 |
|
weight(s) |
= |
3 |
|
|
|
| prec(f) |
= |
2 |
|
weight(f) |
= |
0 |
|
|
|
| prec(g) |
= |
0 |
|
weight(g) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
g(x,s(y)) |
→ |
g(f(x,y),0) |
(2) |
|
g(s(x),y) |
→ |
g(f(x,y),0) |
(3) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.