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 the naturals
| [a] |
= |
18 |
| [g(x1)] |
= |
1 · x1 + 0 |
| [U1(x1, x2, x3)] |
= |
1 · x1 + 8 · x2 + 1 · x3 + 1 |
| [s(x1)] |
= |
8 · x1 + 0 |
| [f(x1, x2)] |
= |
16 · x1 + 3 · x2 + 1 |
| [U2(x1, x2, x3)] |
= |
1 · x1 + 8 · x2 + 2 · x3 + 1 |
| [h(x1)] |
= |
1 · x1 + 0 |
| [c(x1)] |
= |
8 · x1 + 0 |
all of the following rules can be deleted.
|
U1(c(a),x,y) |
→ |
g(s(x)) |
(2) |
|
U2(c(a),x,y) |
→ |
h(s(x)) |
(4) |
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [g(x1)] |
= |
1 · x1 + 1 |
| [U1(x1, x2, x3)] |
= |
2 · x1 + 10 · x2 + 2 · x3 + 23 |
| [s(x1)] |
= |
8 · x1 + 4 |
| [f(x1, x2)] |
= |
14 · x1 + 14 · x2 + 29 |
| [U2(x1, x2, x3)] |
= |
4 · x1 + 6 · x2 + 2 · x3 + 1 |
| [h(x1)] |
= |
1 · x1 + 3 |
| [c(x1)] |
= |
2 · x1 + 1 |
all of the following rules can be deleted.
|
g(s(x)) |
→ |
x |
(5) |
|
h(s(x)) |
→ |
x |
(6) |
1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [g(x1)] |
= |
4 · x1 + 0 |
| [U1(x1, x2, x3)] |
= |
2 · x1 + 19 · x2 + 19 · x3 + 0 |
| [f(x1, x2)] |
= |
27 · x1 + 19 · x2 + 8 |
| [U2(x1, x2, x3)] |
= |
1 · x1 + 11 · x2 + 16 · x3 + 8 |
| [h(x1)] |
= |
16 · x1 + 0 |
| [c(x1)] |
= |
1 · x1 + 0 |
all of the following rules can be deleted.
|
f(x,y) |
→ |
U1(c(g(x)),x,y) |
(1) |
1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1, x2)] |
= |
20 · x1 + 10 · x2 + 16 |
| [U2(x1, x2, x3)] |
= |
4 · x1 + 16 · x2 + 2 · x3 + 15 |
| [h(x1)] |
= |
1 · x1 + 0 |
| [c(x1)] |
= |
1 · x1 + 0 |
all of the following rules can be deleted.
|
f(x,y) |
→ |
U2(c(h(x)),x,y) |
(3) |
1.1.1.1.1.1.1.1 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.