Certification Problem
Input (COPS 408)
The rewrite relation of the following conditional TRS is considered.
|
f(x,y) |
→ |
g(s(x)) |
| c(g(x)) ≈ c(a) |
|
f(x,y) |
→ |
h(s(x)) |
| c(h(x)) ≈ c(a) |
|
g(s(x)) |
→ |
x |
|
h(s(x)) |
→ |
x |
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(x,y)g(s(x))c(g(x))c(a) we get |
| For |
f(x,y)h(s(x))c(h(x))c(a) we get |
| For |
g(s(x))x we get |
| For |
h(s(x))x we get |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the naturals
| [a] |
= |
|
| [g(x1)] |
= |
· x1 +
|
| [U1(x1, x2, x3)] |
= |
· x1 + · x2 + · x3 +
|
| [s(x1)] |
= |
· x1 +
|
| [f(x1, x2)] |
= |
· x1 + · x2 +
|
| [U2(x1, x2, x3)] |
= |
· x1 + · x2 + · x3 +
|
| [h(x1)] |
= |
· x1 +
|
| [c(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
f(x,y) |
→ |
U1(c(g(x)),x,y) |
(1) |
|
U1(c(a),x,y) |
→ |
g(s(x)) |
(2) |
|
f(x,y) |
→ |
U2(c(h(x)),x,y) |
(3) |
|
U2(c(a),x,y) |
→ |
h(s(x)) |
(4) |
|
g(s(x)) |
→ |
x |
(5) |
1.1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(h) |
= |
1 |
|
weight(h) |
= |
2 |
|
|
|
| prec(s) |
= |
0 |
|
weight(s) |
= |
2 |
|
|
|
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
1st
CCP
h(s(y'))
=
g(s(y')) | c(h(y'))c(a)c(g(y'))c(a)
is context-joinable.
-
The
2nd
CCP
g(s(y'))
=
h(s(y')) | c(g(y'))c(a)c(h(y'))c(a)
is context-joinable.