Certification Problem
Input (TPDB TRS_Standard/SK90/4.14)
The rewrite relation of the following TRS is considered.
|
p(s(x)) |
→ |
x |
(1) |
|
s(p(x)) |
→ |
x |
(2) |
|
+(0,y) |
→ |
y |
(3) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(4) |
|
+(p(x),y) |
→ |
p(+(x,y)) |
(5) |
|
minus(0) |
→ |
0 |
(6) |
|
minus(s(x)) |
→ |
p(minus(x)) |
(7) |
|
minus(p(x)) |
→ |
s(minus(x)) |
(8) |
|
*(0,y) |
→ |
0 |
(9) |
|
*(s(x),y) |
→ |
+(*(x,y),y) |
(10) |
|
*(p(x),y) |
→ |
+(*(x,y),minus(y)) |
(11) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(p) |
= |
1 |
|
stat(p) |
= |
mul
|
| prec(s) |
= |
1 |
|
stat(s) |
= |
mul
|
| prec(+) |
= |
2 |
|
stat(+) |
= |
mul
|
| prec(0) |
= |
0 |
|
stat(0) |
= |
mul
|
| prec(*) |
= |
3 |
|
stat(*) |
= |
mul
|
| π(p) |
= |
[1] |
| π(s) |
= |
[1] |
| π(+) |
= |
[1,2] |
| π(0) |
= |
[] |
| π(minus) |
= |
1 |
| π(*) |
= |
[1,2] |
all of the following rules can be deleted.
|
p(s(x)) |
→ |
x |
(1) |
|
s(p(x)) |
→ |
x |
(2) |
|
+(0,y) |
→ |
y |
(3) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(4) |
|
+(p(x),y) |
→ |
p(+(x,y)) |
(5) |
|
*(0,y) |
→ |
0 |
(9) |
|
*(s(x),y) |
→ |
+(*(x,y),y) |
(10) |
|
*(p(x),y) |
→ |
+(*(x,y),minus(y)) |
(11) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(0) |
= |
1 |
|
weight(0) |
= |
1 |
|
|
|
| prec(minus) |
= |
3 |
|
weight(minus) |
= |
2 |
|
|
|
| prec(s) |
= |
0 |
|
weight(s) |
= |
1 |
|
|
|
| prec(p) |
= |
2 |
|
weight(p) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
minus(0) |
→ |
0 |
(6) |
|
minus(s(x)) |
→ |
p(minus(x)) |
(7) |
|
minus(p(x)) |
→ |
s(minus(x)) |
(8) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.