Certification Problem
Input (TPDB TRS_Equational/AProVE_AC_04/AC23)
The rewrite relation of the following equational TRS is considered.
zero(S) |
→ |
S |
(1) |
plus(x,S) |
→ |
x |
(2) |
plus(zero(x),zero(y)) |
→ |
zero(plus(x,y)) |
(3) |
plus(zero(x),un(y)) |
→ |
un(plus(x,y)) |
(4) |
plus(un(x),un(y)) |
→ |
zero(plus(x,plus(y,un(S)))) |
(5) |
times(x,S) |
→ |
S |
(6) |
times(x,times(S,z)) |
→ |
times(S,z) |
(7) |
times(x,zero(y)) |
→ |
zero(times(x,y)) |
(8) |
times(x,times(zero(y),z)) |
→ |
times(zero(times(x,y)),z) |
(9) |
times(x,un(y)) |
→ |
plus(x,zero(times(x,y))) |
(10) |
times(x,times(un(y),z)) |
→ |
times(plus(x,zero(times(x,y))),z) |
(11) |
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 + 1 · x1 · x2
|
[zero(x1)] |
= |
1 · x1
|
[S] |
= |
0 |
[un(x1)] |
= |
2 + 1 · x1
|
the
rules
plus(un(x),un(y)) |
→ |
zero(plus(x,plus(y,un(S)))) |
(5) |
times(x,un(y)) |
→ |
plus(x,zero(times(x,y))) |
(10) |
times(x,times(un(y),z)) |
→ |
times(plus(x,zero(times(x,y))),z) |
(11) |
can be deleted.
1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
[plus(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
[times(x1, x2)] |
= |
2 + 2 · x2 + 2 · x1 + 1 · x1 · x2
|
[zero(x1)] |
= |
2 + 1 · x1
|
[S] |
= |
3 |
[un(x1)] |
= |
2 + 2 · x1
|
the
rules
zero(S) |
→ |
S |
(1) |
plus(x,S) |
→ |
x |
(2) |
plus(zero(x),zero(y)) |
→ |
zero(plus(x,y)) |
(3) |
plus(zero(x),un(y)) |
→ |
un(plus(x,y)) |
(4) |
times(x,S) |
→ |
S |
(6) |
times(x,times(S,z)) |
→ |
times(S,z) |
(7) |
times(x,zero(y)) |
→ |
zero(times(x,y)) |
(8) |
times(x,times(zero(y),z)) |
→ |
times(zero(times(x,y)),z) |
(9) |
can be deleted.
1.1.1 R is empty
There are no rules in the TRS. Hence, it is AC-terminating.