Certification Problem

Input (TPDB TRS_Standard/Various_04/09)

The rewrite relation of the following TRS is considered.

f(x,y,w,w,a) g1(x,x,y,w) (1)
f(x,y,w,a,a) g1(y,x,x,w) (2)
f(x,y,a,a,w) g2(x,y,y,w) (3)
f(x,y,a,w,w) g2(y,y,x,w) (4)
g1(x,x,y,a) h(x,y) (5)
g1(y,x,x,a) h(x,y) (6)
g2(x,y,y,a) h(x,y) (7)
g2(y,y,x,a) h(x,y) (8)
h(x,x) x (9)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f(x1,...,x5)] = 2 + 2 · x1 + 2 · x2 + 2 · x3 + 1 · x4 + 2 · x5
[a] = 0
[g1(x1,...,x4)] = 2 + 1 · x1 + 1 · x2 + 1 · x3 + 2 · x4
[g2(x1,...,x4)] = 2 + 1 · x1 + 1 · x2 + 1 · x3 + 2 · x4
[h(x1, x2)] = 1 + 1 · x1 + 1 · x2
all of the following rules can be deleted.
g1(x,x,y,a) h(x,y) (5)
g1(y,x,x,a) h(x,y) (6)
g2(x,y,y,a) h(x,y) (7)
g2(y,y,x,a) h(x,y) (8)
h(x,x) x (9)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f(x1,...,x5)] = 2 + 2 · x1 + 2 · x2 + 2 · x3 + 2 · x4 + 2 · x5
[a] = 2
[g1(x1,...,x4)] = 1 + 1 · x1 + 1 · x2 + 1 · x3 + 2 · x4
[g2(x1,...,x4)] = 1 + 1 · x1 + 1 · x2 + 1 · x3 + 2 · x4
all of the following rules can be deleted.
f(x,y,w,w,a) g1(x,x,y,w) (1)
f(x,y,w,a,a) g1(y,x,x,w) (2)
f(x,y,a,a,w) g2(x,y,y,w) (3)
f(x,y,a,w,w) g2(y,y,x,w) (4)

1.1.1 R is empty

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