Certification Problem
Input (TPDB TRS_Standard/SK90/2.14)
The rewrite relation of the following TRS is considered.
|
double(0) |
→ |
0 |
(1) |
|
double(s(x)) |
→ |
s(s(double(x))) |
(2) |
|
half(0) |
→ |
0 |
(3) |
|
half(s(0)) |
→ |
0 |
(4) |
|
half(s(s(x))) |
→ |
s(half(x)) |
(5) |
|
-(x,0) |
→ |
x |
(6) |
|
-(s(x),s(y)) |
→ |
-(x,y) |
(7) |
|
if(0,y,z) |
→ |
y |
(8) |
|
if(s(x),y,z) |
→ |
z |
(9) |
|
half(double(x)) |
→ |
x |
(10) |
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
| [double(x1)] |
= |
2 · x1
|
| [0] |
= |
1 |
| [s(x1)] |
= |
2 + 1 · x1
|
| [half(x1)] |
= |
2 + 2 · x1
|
| [-(x1, x2)] |
= |
2 · x1 + 2 · x2
|
| [if(x1, x2, x3)] |
= |
2 · x1 + 1 · x2 + 1 · x3
|
all of the following rules can be deleted.
|
double(0) |
→ |
0 |
(1) |
|
half(0) |
→ |
0 |
(3) |
|
half(s(0)) |
→ |
0 |
(4) |
|
half(s(s(x))) |
→ |
s(half(x)) |
(5) |
|
-(x,0) |
→ |
x |
(6) |
|
-(s(x),s(y)) |
→ |
-(x,y) |
(7) |
|
if(0,y,z) |
→ |
y |
(8) |
|
if(s(x),y,z) |
→ |
z |
(9) |
|
half(double(x)) |
→ |
x |
(10) |
1.1 Rule Removal
Using the
| prec(double) |
= |
1 |
|
stat(double) |
= |
lex
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
| π(double) |
= |
[1] |
| π(s) |
= |
[1] |
all of the following rules can be deleted.
|
double(s(x)) |
→ |
s(s(double(x))) |
(2) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.