Certification Problem
Input (TPDB TRS_Standard/Mixed_TRS/jones4)
The rewrite relation of the following TRS is considered.
|
p(m,n,s(r)) |
→ |
p(m,r,n) |
(1) |
|
p(m,s(n),0) |
→ |
p(0,n,m) |
(2) |
|
p(m,0,0) |
→ |
m |
(3) |
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
| [p(x1, x2, x3)] |
= |
1 · x1 + 2 · x2 + 1 · x3
|
| [s(x1)] |
= |
1 + 2 · x1
|
| [0] |
= |
0 |
all of the following rules can be deleted.
|
p(m,n,s(r)) |
→ |
p(m,r,n) |
(1) |
|
p(m,s(n),0) |
→ |
p(0,n,m) |
(2) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(0) |
= |
1 |
|
weight(0) |
= |
1 |
|
|
|
| prec(p) |
= |
0 |
|
weight(p) |
= |
0 |
|
|
|
all of the following rules can be deleted.
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.