Certification Problem

Input (TPDB TRS_Relative/INVY_15/#3.17a_mset)

The relative rewrite relation R/S is considered where R is the following TRS

app(nil,k) → k (1)
app(l,nil) → l (2)
app(cons(x,l),k) → cons(x,app(l,k)) (3)
sum(cons(x,nil)) → cons(x,nil) (4)
sum(cons(x,cons(y,l))) → sum(cons(plus(x,y),l)) (5)
sum(app(l,cons(x,cons(y,k)))) → sum(app(l,sum(cons(x,cons(y,k))))) (6)
plus(0,y) → y (7)
plus(s(x),y) → s(plus(x,y)) (8)
sum(plus(cons(0,x),cons(y,l))) → pred(sum(cons(s(x),cons(y,l)))) (9)
pred(cons(s(x),nil)) → cons(x,nil) (10)

and S is the following TRS.

cons(x,cons(y,l)) → cons(y,cons(x,l)) (11)

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
[app(x1, x2)] = 1 · x1 + 1 · x2
[nil] = 0
[cons(x1, x2)] = 1 + 1 · x1 + 1 · x2
[sum(x1)] = 1 · x1
[plus(x1, x2)] = 1 + 1 · x1 + 1 · x2
[0] = 1
[s(x1)] = 1 + 1 · x1
[pred(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
plus(0,y) → y (7)
pred(cons(s(x),nil)) → cons(x,nil) (10)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[app(x1, x2)] = 1 + 1 · x1 + 1 · x2
[nil] = 0
[cons(x1, x2)] = 1 + 1 · x1 + 1 · x2
[sum(x1)] = 1 · x1
[plus(x1, x2)] = 1 + 1 · x1 + 1 · x2
[s(x1)] = 1 + 1 · x1
[0] = 1
[pred(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
app(nil,k) → k (1)
app(l,nil) → l (2)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[app(x1, x2)] = 1 · x1 + 1 · x2
[cons(x1, x2)] = 1 + 1 · x1 + 1 · x2
[sum(x1)] = 1 · x1
[nil] = 0
[plus(x1, x2)] = 1 + 1 · x1 + 1 · x2
[s(x1)] = 1 · x1
[0] = 1
[pred(x1)] = 1 · x1
all of the following rules can be deleted.
sum(plus(cons(0,x),cons(y,l))) → pred(sum(cons(s(x),cons(y,l)))) (9)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[app(x1, x2)] = 1 · x1 + 1 · x2
[cons(x1, x2)] = 1 + 1 · x1 + 1 · x2
[sum(x1)] = 1 · x1
[nil] = 0
[plus(x1, x2)] = 1 · x1 + 1 · x2
[s(x1)] = 1 · x1
all of the following rules can be deleted.
sum(cons(x,cons(y,l))) → sum(cons(plus(x,y),l)) (5)

1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[app(x1, x2)] =
0
0
+
1 1
0 0
· x1 +
1 1
0 0
· x2
[cons(x1, x2)] =
0
0
+
1 0
0 0
· x1 +
1 0
0 1
· x2
[sum(x1)] =
0
0
+
1 0
0 0
· x1
[nil] =
0
0
[plus(x1, x2)] =
1
1
+
1 1
1 0
· x1 +
1 1
1 1
· x2
[s(x1)] =
1
1
+
1 0
0 1
· x1
all of the following rules can be deleted.
plus(s(x),y) → s(plus(x,y)) (8)

1.1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[app(x1, x2)] =
0
0
+
1 1
0 1
· x1 +
1 0
0 0
· x2
[cons(x1, x2)] =
0
1
+
1 0
0 0
· x1 +
1 0
0 1
· x2
[sum(x1)] =
0
1
+
1 0
0 0
· x1
[nil] =
0
0
all of the following rules can be deleted.
app(cons(x,l),k) → cons(x,app(l,k)) (3)

1.1.1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[sum(x1)] =
1
1
+
1 0
0 0
· x1
[cons(x1, x2)] =
0
1
+
1 0
0 0
· x1 +
1 0
0 1
· x2
[nil] =
0
0
[app(x1, x2)] =
0
0
+
1 0
0 0
· x1 +
1 1
0 0
· x2
all of the following rules can be deleted.
sum(cons(x,nil)) → cons(x,nil) (4)

1.1.1.1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[sum(x1)] =
0
0
+
1 0
0 0
· x1
[app(x1, x2)] =
0
0
+
1 0
0 0
· x1 +
1 1
0 0
· x2
[cons(x1, x2)] =
0
1
+
1 0
0 0
· x1 +
1 0
0 0
· x2
all of the following rules can be deleted.
sum(app(l,cons(x,cons(y,k)))) → sum(app(l,sum(cons(x,cons(y,k))))) (6)

1.1.1.1.1.1.1.1.1 R is empty

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