Certification Problem
Input (TPDB TRS_Standard/TCT_12/polycounter-5)
The rewrite relation of the following TRS is considered.
|
f(s(x1),x2,x3,x4,x5) |
→ |
f(x1,x2,x3,x4,x5) |
(1) |
|
f(0,s(x2),x3,x4,x5) |
→ |
f(x2,x2,x3,x4,x5) |
(2) |
|
f(0,0,s(x3),x4,x5) |
→ |
f(x3,x3,x3,x4,x5) |
(3) |
|
f(0,0,0,s(x4),x5) |
→ |
f(x4,x4,x4,x4,x5) |
(4) |
|
f(0,0,0,0,s(x5)) |
→ |
f(x5,x5,x5,x5,x5) |
(5) |
|
f(0,0,0,0,0) |
→ |
0 |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1,...,x5)] |
= |
4 · x1 + 2 · x2 + 3 · x3 + 3 · x4 + 18 · x5 + 8 |
| [s(x1)] |
= |
4 · x1 + 1 |
| [0] |
= |
0 |
all of the following rules can be deleted.
|
f(s(x1),x2,x3,x4,x5) |
→ |
f(x1,x2,x3,x4,x5) |
(1) |
|
f(0,s(x2),x3,x4,x5) |
→ |
f(x2,x2,x3,x4,x5) |
(2) |
|
f(0,0,s(x3),x4,x5) |
→ |
f(x3,x3,x3,x4,x5) |
(3) |
|
f(0,0,0,s(x4),x5) |
→ |
f(x4,x4,x4,x4,x5) |
(4) |
|
f(0,0,0,0,s(x5)) |
→ |
f(x5,x5,x5,x5,x5) |
(5) |
|
f(0,0,0,0,0) |
→ |
0 |
(6) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.