Certification Problem
Input (TPDB TRS_Standard/SK90/4.53)
The rewrite relation of the following TRS is considered.
|
f(a) |
→ |
b |
(1) |
|
f(c) |
→ |
d |
(2) |
|
f(g(x,y)) |
→ |
g(f(x),f(y)) |
(3) |
|
f(h(x,y)) |
→ |
g(h(y,f(x)),h(x,f(y))) |
(4) |
|
g(x,x) |
→ |
h(e,x) |
(5) |
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(e) |
= |
0 |
|
status(e) |
= |
[] |
|
list-extension(e) |
= |
Lex |
| prec(h) |
= |
0 |
|
status(h) |
= |
[1, 2] |
|
list-extension(h) |
= |
Lex |
| prec(g) |
= |
0 |
|
status(g) |
= |
[2, 1] |
|
list-extension(g) |
= |
Lex |
| prec(d) |
= |
0 |
|
status(d) |
= |
[] |
|
list-extension(d) |
= |
Lex |
| prec(c) |
= |
0 |
|
status(c) |
= |
[] |
|
list-extension(c) |
= |
Lex |
| prec(b) |
= |
0 |
|
status(b) |
= |
[] |
|
list-extension(b) |
= |
Lex |
| prec(f) |
= |
3 |
|
status(f) |
= |
[1] |
|
list-extension(f) |
= |
Lex |
| prec(a) |
= |
0 |
|
status(a) |
= |
[] |
|
list-extension(a) |
= |
Lex |
and the following
Max-polynomial interpretation
| [e] |
=
|
max(0) |
| [h(x1, x2)] |
=
|
max(0, 2 + 1 · x1, 0 + 1 · x2) |
| [g(x1, x2)] |
=
|
max(4, 1 + 1 · x1, 0 + 1 · x2) |
| [d] |
=
|
max(0) |
| [c] |
=
|
max(4) |
| [b] |
=
|
max(0) |
| [f(x1)] |
=
|
max(0, 4 + 1 · x1) |
| [a] |
=
|
max(4) |
all of the following rules can be deleted.
|
f(a) |
→ |
b |
(1) |
|
f(c) |
→ |
d |
(2) |
|
f(g(x,y)) |
→ |
g(f(x),f(y)) |
(3) |
|
f(h(x,y)) |
→ |
g(h(y,f(x)),h(x,f(y))) |
(4) |
|
g(x,x) |
→ |
h(e,x) |
(5) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.