Certification Problem

Input (COPS 340)

The rewrite relation of the following conditional TRS is considered.

f(x',x'') g(x) | x'x, x''x
f(y',h(y'')) g(y) | y'y, y''y

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',x'')g(x)x'xx''x we get
For f(y',h(y''))g(y)y'yy''y we get

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[h(x1)] = 16 · x1 + 0
[U1(x1, x2, x3)] = 5 · x1 + 25 · x2 + 6 · x3 + 0
[g(x1)] = 7 · x1 + 0
[U3(x1, x2, x3)] = 10 · x1 + 20 · x2 + 8 · x3 + 6
[f(x1, x2)] = 30 · x1 + 6 · x2 + 6
[U4(x1,...,x4)] = 3 · x1 + 1 · x2 + 2 · x3 + 10 · x4 + 6
[U2(x1,...,x4)] = 2 · x1 + 16 · x2 + 4 · x3 + 5 · x4 + 0
all of the following rules can be deleted.
f(x',x'') U1(x',x',x'') (1)
U4(y,y',y'',y) g(y) (6)

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[h(x1)] =
4 0
6 2
· x1 +
4 0
4 0
[U1(x1, x2, x3)] =
4 0
0 0
· x1 +
2 0
0 0
· x2 +
5 6
1 4
· x3 +
4 0
4 0
[g(x1)] =
6 6
0 0
· x1 +
0 0
0 0
[U3(x1, x2, x3)] =
1 0
0 0
· x1 +
6 4
0 0
· x2 +
7 0
0 2
· x3 +
0 0
0 0
[f(x1, x2)] =
7 5
4 1
· x1 +
1 1
0 1
· x2 +
3 0
1 0
[U4(x1,...,x4)] =
4 0
0 0
· x1 +
6 4
0 0
· x2 +
3 0
0 2
· x3 +
1 0
0 0
· x4 +
0 0
0 0
[U2(x1,...,x4)] =
4 6
0 0
· x1 +
2 0
0 0
· x2 +
1 0
0 0
· x3 +
2 0
0 0
· x4 +
0 0
4 0
all of the following rules can be deleted.
U1(x,x',x'') U2(x'',x',x'',x) (2)
f(y',h(y'')) U3(y',y',y'') (4)

1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[g(x1)] =
2 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[U3(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
2 1 1
0 0 2
2 1 3
· x3 +
1 0 0
0 0 0
0 0 0
[U4(x1,...,x4)] =
1 0 0
0 0 0
1 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
0 0 2
1 0 2
· x3 +
1 0 0
0 0 0
0 0 0
· x4 +
0 0 0
0 0 0
0 0 0
[U2(x1,...,x4)] =
1 1 1
2 0 2
0 3 0
· x1 +
2 0 0
0 0 0
0 0 0
· x2 +
2 0 0
0 0 0
0 0 0
· x3 +
1 1 1
0 0 0
1 1 0
· x4 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
U3(y,y',y'') U4(y'',y',y'',y) (5)

1.1.1.1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(g) = 0 weight(g) = 3
prec(U2) = 1 weight(U2) = 0
all of the following rules can be deleted.
U2(x,x',x'',x) g(x) (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.