Certification Problem
Input (TPDB TRS_Standard/AG01/#3.12)
The rewrite relation of the following TRS is considered.
|
app(nil,y) |
→ |
y |
(1) |
|
app(add(n,x),y) |
→ |
add(n,app(x,y)) |
(2) |
|
reverse(nil) |
→ |
nil |
(3) |
|
reverse(add(n,x)) |
→ |
app(reverse(x),add(n,nil)) |
(4) |
|
shuffle(nil) |
→ |
nil |
(5) |
|
shuffle(add(n,x)) |
→ |
add(n,shuffle(reverse(x))) |
(6) |
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
| [app(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [nil] |
= |
0 |
| [add(x1, x2)] |
= |
2 + 2 · x1 + 1 · x2
|
| [reverse(x1)] |
= |
1 · x1
|
| [shuffle(x1)] |
= |
2 + 2 · x1
|
all of the following rules can be deleted.
|
shuffle(nil) |
→ |
nil |
(5) |
|
shuffle(add(n,x)) |
→ |
add(n,shuffle(reverse(x))) |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [app(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [nil] |
= |
0 |
| [add(x1, x2)] |
= |
2 · x1 + 1 · x2
|
| [reverse(x1)] |
= |
2 + 1 · x1
|
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [app(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [nil] |
= |
0 |
| [add(x1, x2)] |
= |
1 + 2 · x1 + 1 · x2
|
| [reverse(x1)] |
= |
2 · x1
|
all of the following rules can be deleted.
|
reverse(add(n,x)) |
→ |
app(reverse(x),add(n,nil)) |
(4) |
1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(nil) |
= |
0 |
|
weight(nil) |
= |
1 |
|
|
|
| prec(app) |
= |
2 |
|
weight(app) |
= |
0 |
|
|
|
| prec(add) |
= |
1 |
|
weight(add) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
app(nil,y) |
→ |
y |
(1) |
|
app(add(n,x),y) |
→ |
add(n,app(x,y)) |
(2) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.