Certification Problem
Input (TPDB TRS_Standard/SK90/2.07)
The rewrite relation of the following TRS is considered.
|
f(0,y) |
→ |
y |
(1) |
|
f(x,0) |
→ |
x |
(2) |
|
f(i(x),y) |
→ |
i(x) |
(3) |
|
f(f(x,y),z) |
→ |
f(x,f(y,z)) |
(4) |
|
f(g(x,y),z) |
→ |
g(f(x,z),f(y,z)) |
(5) |
|
f(1,g(x,y)) |
→ |
x |
(6) |
|
f(2,g(x,y)) |
→ |
y |
(7) |
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(2) |
= |
0 |
|
status(2) |
= |
[] |
|
list-extension(2) |
= |
Lex |
| prec(1) |
= |
0 |
|
status(1) |
= |
[] |
|
list-extension(1) |
= |
Lex |
| prec(g) |
= |
0 |
|
status(g) |
= |
[1, 2] |
|
list-extension(g) |
= |
Lex |
| prec(i) |
= |
0 |
|
status(i) |
= |
[1] |
|
list-extension(i) |
= |
Lex |
| prec(f) |
= |
4 |
|
status(f) |
= |
[1, 2] |
|
list-extension(f) |
= |
Lex |
| prec(0) |
= |
0 |
|
status(0) |
= |
[] |
|
list-extension(0) |
= |
Lex |
and the following
Max-polynomial interpretation
| [2] |
=
|
max(0) |
| [1] |
=
|
max(0) |
| [g(x1, x2)] |
=
|
max(5, 0 + 1 · x1, 0 + 1 · x2) |
| [i(x1)] |
=
|
7 + 1 · x1
|
| [f(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
| [0] |
=
|
max(0) |
all of the following rules can be deleted.
|
f(0,y) |
→ |
y |
(1) |
|
f(x,0) |
→ |
x |
(2) |
|
f(i(x),y) |
→ |
i(x) |
(3) |
|
f(f(x,y),z) |
→ |
f(x,f(y,z)) |
(4) |
|
f(g(x,y),z) |
→ |
g(f(x,z),f(y,z)) |
(5) |
|
f(1,g(x,y)) |
→ |
x |
(6) |
|
f(2,g(x,y)) |
→ |
y |
(7) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.