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 ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(f) |
= |
3 |
|
status(f) |
= |
[1, 2] |
|
list-extension(f) |
= |
Lex |
| prec(+) |
= |
1 |
|
status(+) |
= |
[1, 2] |
|
list-extension(+) |
= |
Lex |
| prec(b) |
= |
0 |
|
status(b) |
= |
[] |
|
list-extension(b) |
= |
Lex |
| prec(a) |
= |
0 |
|
status(a) |
= |
[] |
|
list-extension(a) |
= |
Lex |
and the following
Max-polynomial interpretation
| [f(x1, x2)] |
=
|
max(4, 2 + 1 · x1, 0 + 1 · x2) |
| [+(x1, x2)] |
=
|
max(4, 0 + 1 · x1, 0 + 1 · x2) |
| [b] |
=
|
max(0) |
| [a] |
=
|
6 |
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.