Certification Problem
Input (TPDB TRS_Equational/AProVE_AC_04/AC17)
The rewrite relation of the following equational TRS is considered.
|
0(S) |
→ |
S |
(1) |
|
plus(x,S) |
→ |
x |
(2) |
|
plus(0(x),0(y)) |
→ |
0(plus(x,y)) |
(3) |
|
plus(0(x),1(y)) |
→ |
1(plus(x,y)) |
(4) |
|
plus(1(x),1(y)) |
→ |
0(1(plus(plus(x,y),S))) |
(5) |
|
times(x,S) |
→ |
S |
(6) |
|
times(x,0(y)) |
→ |
0(times(x,y)) |
(7) |
|
times(x,1(y)) |
→ |
plus(x,0(times(x,y))) |
(8) |
Associative symbols: plus, times
Commutative symbols: plus, times
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
| [plus(x1, x2)] |
= |
1 · x2 + 1 · x1
|
| [times(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [0(x1)] |
= |
2 + 1 · x1
|
| [S] |
= |
0 |
| [1(x1)] |
= |
2 + 1 · x1
|
the
rules
|
0(S) |
→ |
S |
(1) |
|
plus(0(x),0(y)) |
→ |
0(plus(x,y)) |
(3) |
|
plus(0(x),1(y)) |
→ |
1(plus(x,y)) |
(4) |
|
times(x,S) |
→ |
S |
(6) |
|
times(x,0(y)) |
→ |
0(times(x,y)) |
(7) |
|
times(x,1(y)) |
→ |
plus(x,0(times(x,y))) |
(8) |
can be deleted.
1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
1 · x2 + 1 · x1
|
| [times(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [S] |
= |
2 |
| [1(x1)] |
= |
2 + 1 · x1
|
| [0(x1)] |
= |
1 · x1
|
the
rule
can be deleted.
1.1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
1 · x2 + 1 · x1
|
| [times(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [1(x1)] |
= |
2 + 1 · x1
|
| [0(x1)] |
= |
1 · x1
|
| [S] |
= |
0 |
the
rule
|
plus(1(x),1(y)) |
→ |
0(1(plus(plus(x,y),S))) |
(5) |
can be deleted.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is AC-terminating.