Certification Problem
Input (TPDB TRS_Standard/HirokawaMiddeldorp_04/t007)
The rewrite relation of the following TRS is considered.
|
f(a) |
→ |
f(b) |
(1) |
|
g(b) |
→ |
g(a) |
(2) |
|
f(x) |
→ |
g(x) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Closure Under Flat Contexts
Using the flat contexts
{f(☐), g(☐)}
We obtain the transformed TRS
|
f(a) |
→ |
f(b) |
(1) |
|
g(b) |
→ |
g(a) |
(2) |
|
f(f(x)) |
→ |
f(g(x)) |
(4) |
|
g(f(x)) |
→ |
g(g(x)) |
(5) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
fa(a) |
→ |
fb(b) |
(6) |
|
gb(b) |
→ |
ga(a) |
(7) |
|
ff(ff(x)) |
→ |
fg(gf(x)) |
(8) |
|
ff(fa(x)) |
→ |
fg(ga(x)) |
(9) |
|
ff(fb(x)) |
→ |
fg(gb(x)) |
(10) |
|
ff(fg(x)) |
→ |
fg(gg(x)) |
(11) |
|
gf(ff(x)) |
→ |
gg(gf(x)) |
(12) |
|
gf(fa(x)) |
→ |
gg(ga(x)) |
(13) |
|
gf(fb(x)) |
→ |
gg(gb(x)) |
(14) |
|
gf(fg(x)) |
→ |
gg(gg(x)) |
(15) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [fa(x1)] |
= |
1 · x1 + 2 |
| [a] |
= |
0 |
| [fb(x1)] |
= |
1 · x1
|
| [b] |
= |
1 |
| [gb(x1)] |
= |
1 · x1
|
| [ga(x1)] |
= |
1 · x1
|
| [ff(x1)] |
= |
1 · x1 + 1 |
| [fg(x1)] |
= |
1 · x1
|
| [gf(x1)] |
= |
1 · x1 + 1 |
| [gg(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
fa(a) |
→ |
fb(b) |
(6) |
|
gb(b) |
→ |
ga(a) |
(7) |
|
ff(ff(x)) |
→ |
fg(gf(x)) |
(8) |
|
ff(fa(x)) |
→ |
fg(ga(x)) |
(9) |
|
ff(fb(x)) |
→ |
fg(gb(x)) |
(10) |
|
ff(fg(x)) |
→ |
fg(gg(x)) |
(11) |
|
gf(ff(x)) |
→ |
gg(gf(x)) |
(12) |
|
gf(fa(x)) |
→ |
gg(ga(x)) |
(13) |
|
gf(fb(x)) |
→ |
gg(gb(x)) |
(14) |
|
gf(fg(x)) |
→ |
gg(gg(x)) |
(15) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.