Certification Problem
Input (TPDB TRS_Standard/SK90/2.08)
The rewrite relation of the following TRS is considered.
|
+(a,b) |
→ |
+(b,a) |
(1) |
|
+(a,+(b,z)) |
→ |
+(b,+(a,z)) |
(2) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(3) |
|
f(a,y) |
→ |
a |
(4) |
|
f(b,y) |
→ |
b |
(5) |
|
f(+(x,y),z) |
→ |
+(f(x,z),f(y,z)) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(+) |
= |
1 |
|
stat(+) |
= |
lex
|
| prec(a) |
= |
1 |
|
stat(a) |
= |
mul
|
| prec(b) |
= |
0 |
|
stat(b) |
= |
mul
|
| prec(f) |
= |
2 |
|
stat(f) |
= |
mul
|
| π(+) |
= |
[1,2] |
| π(a) |
= |
[] |
| π(b) |
= |
[] |
| π(f) |
= |
[1,2] |
all of the following rules can be deleted.
|
+(a,b) |
→ |
+(b,a) |
(1) |
|
+(a,+(b,z)) |
→ |
+(b,+(a,z)) |
(2) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(3) |
|
f(a,y) |
→ |
a |
(4) |
|
f(b,y) |
→ |
b |
(5) |
|
f(+(x,y),z) |
→ |
+(f(x,z),f(y,z)) |
(6) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.