Certification Problem
Input (TPDB TRS_Standard/SK90/2.02)
The rewrite relation of the following TRS is considered.
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(1) |
|
+(f(x),f(y)) |
→ |
f(+(x,y)) |
(2) |
|
+(f(x),+(f(y),z)) |
→ |
+(f(+(x,y)),z) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(f) |
= |
1 |
|
weight(f) |
= |
1 |
|
|
|
| prec(+) |
= |
0 |
|
weight(+) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(1) |
|
+(f(x),f(y)) |
→ |
f(+(x,y)) |
(2) |
|
+(f(x),+(f(y),z)) |
→ |
+(f(+(x,y)),z) |
(3) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.