Certification Problem
Input (TPDB TRS_Standard/Zantema_05/z09)
The rewrite relation of the following TRS is considered.
|
a(lambda(x),y) |
→ |
lambda(a(x,p(1,a(y,t)))) |
(1) |
|
a(p(x,y),z) |
→ |
p(a(x,z),a(y,z)) |
(2) |
|
a(a(x,y),z) |
→ |
a(x,a(y,z)) |
(3) |
|
lambda(x) |
→ |
x |
(4) |
|
a(x,y) |
→ |
x |
(5) |
|
a(x,y) |
→ |
y |
(6) |
|
p(x,y) |
→ |
x |
(7) |
|
p(x,y) |
→ |
y |
(8) |
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(p) |
= |
0 |
|
status(p) |
= |
[1, 2] |
|
list-extension(p) |
= |
Lex |
| prec(t) |
= |
0 |
|
status(t) |
= |
[] |
|
list-extension(t) |
= |
Lex |
| prec(1) |
= |
0 |
|
status(1) |
= |
[] |
|
list-extension(1) |
= |
Lex |
| prec(a) |
= |
1 |
|
status(a) |
= |
[1, 2] |
|
list-extension(a) |
= |
Lex |
| prec(lambda) |
= |
0 |
|
status(lambda) |
= |
[1] |
|
list-extension(lambda) |
= |
Lex |
and the following
Max-polynomial interpretation
| [p(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
| [t] |
=
|
max(0) |
| [1] |
=
|
max(0) |
| [a(x1, x2)] |
=
|
0 + 1 · x1 + 1 · x2
|
| [lambda(x1)] |
=
|
2 + 1 · x1
|
all of the following rules can be deleted.
|
a(lambda(x),y) |
→ |
lambda(a(x,p(1,a(y,t)))) |
(1) |
|
a(p(x,y),z) |
→ |
p(a(x,z),a(y,z)) |
(2) |
|
a(a(x,y),z) |
→ |
a(x,a(y,z)) |
(3) |
|
lambda(x) |
→ |
x |
(4) |
|
a(x,y) |
→ |
x |
(5) |
|
a(x,y) |
→ |
y |
(6) |
|
p(x,y) |
→ |
x |
(7) |
|
p(x,y) |
→ |
y |
(8) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.