Certification Problem
Input (TPDB TRS_Relative/Mixed_relative_TRS/relsubst)
The relative rewrite relation R/S is considered where R is the following TRS
o(lambda(x),y) |
→ |
lambda(o(x,d(1,o(y,p)))) |
(1) |
o(d(x,y),z) |
→ |
d(o(x,z),o(y,z)) |
(2) |
o(o(x,y),z) |
→ |
o(x,o(y,z)) |
(3) |
lambda(x) |
→ |
x |
(4) |
o(x,y) |
→ |
x |
(5) |
o(x,y) |
→ |
y |
(6) |
d(x,y) |
→ |
x |
(7) |
d(x,y) |
→ |
y |
(8) |
and S is the following TRS.
o(x,y) |
→ |
d(x,y) |
(9) |
o(x,y) |
→ |
d(y,x) |
(10) |
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(d) |
= |
0 |
|
status(d) |
= |
[1, 2] |
|
list-extension(d) |
= |
Lex |
prec(p) |
= |
0 |
|
status(p) |
= |
[] |
|
list-extension(p) |
= |
Lex |
prec(1) |
= |
0 |
|
status(1) |
= |
[] |
|
list-extension(1) |
= |
Lex |
prec(o) |
= |
1 |
|
status(o) |
= |
[1, 2] |
|
list-extension(o) |
= |
Lex |
prec(lambda) |
= |
0 |
|
status(lambda) |
= |
[1] |
|
list-extension(lambda) |
= |
Lex |
and the following
Max-polynomial interpretation
[d(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
[p] |
=
|
0 |
[1] |
=
|
0 |
[o(x1, x2)] |
=
|
0 + 1 · x1 + 1 · x2
|
[lambda(x1)] |
=
|
max(4, 4 + 1 · x1) |
all of the following rules can be deleted.
o(lambda(x),y) |
→ |
lambda(o(x,d(1,o(y,p)))) |
(1) |
o(d(x,y),z) |
→ |
d(o(x,z),o(y,z)) |
(2) |
o(o(x,y),z) |
→ |
o(x,o(y,z)) |
(3) |
lambda(x) |
→ |
x |
(4) |
o(x,y) |
→ |
x |
(5) |
o(x,y) |
→ |
y |
(6) |
d(x,y) |
→ |
x |
(7) |
d(x,y) |
→ |
y |
(8) |
o(x,y) |
→ |
d(x,y) |
(9) |
o(x,y) |
→ |
d(y,x) |
(10) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.