Certification Problem
Input (COPS 491)
The rewrite relation of the following conditional TRS is considered.
|
f(g(x)) |
→ |
b |
| x ≈ a
|
|
g(x) |
→ |
c |
| x ≈ c
|
Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by ConCon @ CoCo 2020)
1 Quasi-reductive SDTRS where all CCPs are joinable
The given strongly deterministic oriented 3-CTRS is quasi-reductive and all CCPs are joinable.
1.1 Quasi-Reductive CTRS
The given CTRS is quasi-reductive
1.1.1 Unraveling
To prove that the CTRS is quasi-reductive, we show termination of the following
unraveled system.
| For |
f(g(x))bxa we get |
| For |
g(x)cxc we get |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [b] |
= |
0 |
| [f(x1)] |
= |
2 · x1 + 0 |
| [a] |
= |
0 |
| [U2(x1, x2)] |
= |
1 · x1 + 2 · x2 + 4 |
| [g(x1)] |
= |
16 · x1 + 20 |
| [c] |
= |
4 |
| [U1(x1, x2)] |
= |
30 · x1 + 2 · x2 + 0 |
all of the following rules can be deleted.
|
f(g(x)) |
→ |
U1(x,x) |
(1) |
|
g(x) |
→ |
U2(x,x) |
(3) |
|
U2(c,x) |
→ |
c |
(4) |
1.1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(b) |
= |
0 |
|
weight(b) |
= |
2 |
|
|
|
| prec(a) |
= |
0 |
|
weight(a) |
= |
1 |
|
|
|
| prec(U1) |
= |
1 |
|
weight(U1) |
= |
0 |
|
|
|
all of the following rules can be deleted.
1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.
1.2 All CCPs are joinable
A CCP is joinable if it is context-joinable, infeasible, or unfeasible.
-
The
2nd
CCP stemming from the overlap of rules
and
with mgu
{y/z}
is unfeasible.