Certification Problem

Input (COPS 316)

The rewrite relation of the following conditional TRS is considered.

f(x,y) x | g(x) ≈ z, g(y) ≈ z
g(x) c | dc

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by ConCon @ CoCo 2020)

1 Removal of Infeasible Rules

We may safely remove rules with infeasible conditions. They do not influence the rewrite relation in any way.

1.1 Rules with Infeasible Conditions

1.2 Quasi-reductive SDTRS where all CCPs are joinable

The given strongly deterministic oriented 3-CTRS is quasi-reductive and all CCPs are joinable.

1.2.1 Quasi-Reductive CTRS

The given CTRS is quasi-reductive

1.2.1.1 Unraveling

To prove that the CTRS is quasi-reductive, we show termination of the following unraveled system.

For f(x,y)xg(x)zg(y)z we get

1.2.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[g(x1)] = 2 · x1 + 0
[U2(x1,...,x4)] = 4 · x1 + 2 · x2 + 1 · x3 + 6 · x4 + 8
[f(x1, x2)] = 26 · x1 + 9 · x2 + 16
[U1(x1, x2, x3)] = 12 · x1 + 2 · x2 + 9 · x3 + 16
all of the following rules can be deleted.
U1(z,x,y) U2(g(y),x,y,z) (3)
U2(z,x,y,z) x (4)

1.2.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[g(x1)] = 18 · x1 + 0
[f(x1, x2)] = 27 · x1 + 18 · x2 + 1
[U1(x1, x2, x3)] = 1 · x1 + 9 · x2 + 2 · x3 + 0
all of the following rules can be deleted.
f(x,y) U1(g(x),x,y) (2)

1.2.1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.

1.2.2 All CCPs are joinable

A CCP is joinable if it is context-joinable, infeasible, or unfeasible.