Certification Problem
Input (TPDB TRS_Equational/AProVE_AC_04/AC06)
The rewrite relation of the following equational TRS is considered.
|
s(p(x)) |
→ |
x |
(1) |
|
p(s(x)) |
→ |
x |
(2) |
|
plus(0,y) |
→ |
y |
(3) |
|
plus(s(x),y) |
→ |
s(plus(x,y)) |
(4) |
|
plus(p(x),y) |
→ |
p(plus(x,y)) |
(5) |
|
plus(i(x),x) |
→ |
0 |
(6) |
|
plus(x,plus(i(x),y)) |
→ |
y |
(7) |
|
i(0) |
→ |
0 |
(8) |
|
i(s(x)) |
→ |
p(i(x)) |
(9) |
|
i(p(x)) |
→ |
s(i(x)) |
(10) |
|
i(i(x)) |
→ |
x |
(11) |
|
i(plus(x,y)) |
→ |
plus(i(y),i(x)) |
(12) |
|
times(0,y) |
→ |
0 |
(13) |
|
times(s(x),y) |
→ |
plus(times(x,y),y) |
(14) |
|
times(p(x),y) |
→ |
plus(times(x,y),i(y)) |
(15) |
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)] |
= |
1 · x2 + 1 · x1 + 2 · x1 · x2
|
| [s(x1)] |
= |
2 + 1 · x1
|
| [p(x1)] |
= |
2 + 1 · x1
|
| [0] |
= |
0 |
| [i(x1)] |
= |
1 · x1
|
the
rules
|
s(p(x)) |
→ |
x |
(1) |
|
p(s(x)) |
→ |
x |
(2) |
|
times(s(x),y) |
→ |
plus(times(x,y),y) |
(14) |
|
times(p(x),y) |
→ |
plus(times(x,y),i(y)) |
(15) |
can be deleted.
1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
1 · x2 + 1 · x1 + 2 · x1 · x2
|
| [times(x1, x2)] |
= |
2 + 1 · x2 + 1 · x1
|
| [0] |
= |
0 |
| [s(x1)] |
= |
2 + 2 · x1
|
| [p(x1)] |
= |
2 + 2 · x1
|
| [i(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 + 2 · x1 · x2
|
| [times(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [0] |
= |
0 |
| [s(x1)] |
= |
3 + 3 · x1
|
| [p(x1)] |
= |
2 + 2 · x1
|
| [i(x1)] |
= |
1 · x1 + 2 · x1 · x1
|
the
rules
|
i(s(x)) |
→ |
p(i(x)) |
(9) |
|
i(p(x)) |
→ |
s(i(x)) |
(10) |
can be deleted.
1.1.1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
2 + 1 · x2 + 1 · x1
|
| [times(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [0] |
= |
0 |
| [s(x1)] |
= |
1 · x1
|
| [p(x1)] |
= |
1 · x1
|
| [i(x1)] |
= |
2 · x1 + 2 · x1 · x1
|
the
rules
|
plus(0,y) |
→ |
y |
(3) |
|
plus(i(x),x) |
→ |
0 |
(6) |
|
plus(x,plus(i(x),y)) |
→ |
y |
(7) |
|
i(plus(x,y)) |
→ |
plus(i(y),i(x)) |
(12) |
can be deleted.
1.1.1.1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
|
| [times(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [s(x1)] |
= |
2 + 2 · x1
|
| [p(x1)] |
= |
1 · x1
|
| [i(x1)] |
= |
1 + 1 · x1
|
| [0] |
= |
3 |
the
rules
|
plus(s(x),y) |
→ |
s(plus(x,y)) |
(4) |
|
i(0) |
→ |
0 |
(8) |
|
i(i(x)) |
→ |
x |
(11) |
can be deleted.
1.1.1.1.1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
|
| [times(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [p(x1)] |
= |
2 + 1 · x1
|
the
rule
|
plus(p(x),y) |
→ |
p(plus(x,y)) |
(5) |
can be deleted.
1.1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is AC-terminating.