Certification Problem
Input (TPDB TRS_Equational/Mixed_AC/differ)
The rewrite relation of the following equational TRS is considered.
|
dx(X) |
→ |
1 |
(1) |
|
dx(0) |
→ |
0 |
(2) |
|
dx(1) |
→ |
0 |
(3) |
|
dx(a) |
→ |
0 |
(4) |
|
dx(+(f,g)) |
→ |
+(dx(f),dx(g)) |
(5) |
|
dx(*(f,g)) |
→ |
+(*(dx(f),g),*(dx(g),f)) |
(6) |
|
dx(-(f,g)) |
→ |
-(dx(f),dx(g)) |
(7) |
|
dx(neg(f)) |
→ |
neg(dx(f)) |
(8) |
|
dx(/(f,g)) |
→ |
-(/(dx(f),g),/(*(dx(g),f),exp(g,2))) |
(9) |
|
dx(ln(f)) |
→ |
/(dx(f),f) |
(10) |
|
dx(exp(f,g)) |
→ |
+(*(dx(f),*(exp(f,-(g,1)),g)),*(dx(g),*(exp(f,g),ln(f)))) |
(11) |
Associative symbols: *, +
Commutative symbols: *, +
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [*(x1, x2)] |
= |
2 + 1 · x2 + 1 · x1
|
| [+(x1, x2)] |
= |
1 · x2 + 1 · x1
|
| [dx(x1)] |
= |
2 · x1 + 3 · x1 · x1
|
| [X] |
= |
1 |
| [1] |
= |
0 |
| [0] |
= |
0 |
| [a] |
= |
0 |
| [-(x1, x2)] |
= |
3 + 2 · x2 + 1 · x1
|
| [neg(x1)] |
= |
2 · x1
|
| [/(x1, x2)] |
= |
3 + 2 · x2 + 2 · x1
|
| [exp(x1, x2)] |
= |
2 + 1 · x2 + 2 · x1
|
| [2] |
= |
1 |
| [ln(x1)] |
= |
1 + 1 · x1 + 3 · x1 · x1
|
the
rules
|
dx(X) |
→ |
1 |
(1) |
|
dx(*(f,g)) |
→ |
+(*(dx(f),g),*(dx(g),f)) |
(6) |
|
dx(-(f,g)) |
→ |
-(dx(f),dx(g)) |
(7) |
|
dx(/(f,g)) |
→ |
-(/(dx(f),g),/(*(dx(g),f),exp(g,2))) |
(9) |
|
dx(ln(f)) |
→ |
/(dx(f),f) |
(10) |
can be deleted.
1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [*(x1, x2)] |
= |
1 · x2 + 1 · x1
|
| [+(x1, x2)] |
= |
2 + 1 · x2 + 1 · x1
|
| [dx(x1)] |
= |
3 + 3 · x1 · x1
|
| [0] |
= |
1 |
| [1] |
= |
0 |
| [a] |
= |
1 |
| [neg(x1)] |
= |
3 + 3 · x1
|
| [exp(x1, x2)] |
= |
2 + 2 · x2 + 1 · x1 + 1 · x1 · x2
|
| [-(x1, x2)] |
= |
1 · x2 + 1 · x1
|
| [ln(x1)] |
= |
1 · x1
|
the
rules
|
dx(0) |
→ |
0 |
(2) |
|
dx(1) |
→ |
0 |
(3) |
|
dx(a) |
→ |
0 |
(4) |
|
dx(+(f,g)) |
→ |
+(dx(f),dx(g)) |
(5) |
|
dx(neg(f)) |
→ |
neg(dx(f)) |
(8) |
|
dx(exp(f,g)) |
→ |
+(*(dx(f),*(exp(f,-(g,1)),g)),*(dx(g),*(exp(f,g),ln(f)))) |
(11) |
can be deleted.
1.1.1 R is empty
There are no rules in the TRS. Hence, it is AC-terminating.