Certification Problem

Input (TPDB TRS_Equational/Mixed_AC_and_C/AC29)

The rewrite relation of the following equational TRS is considered.

1 s(0) (1)
2 s(1) (2)
3 s(2) (3)
4 s(3) (4)
5 s(4) (5)
6 s(5) (6)
7 s(6) (7)
8 s(7) (8)
9 s(8) (9)
max'(0,x) x (10)
max'(s(x),s(y)) s(max'(x,y)) (11)
app(empty,X) X (12)
max(singl(x)) x (13)
max(app(singl(x),Y)) max2(x,Y) (14)
max2(x,empty) x (15)
max2(x,singl(y)) max'(x,y) (16)
max2(x,app(singl(y),Z)) max2(max'(x,y),Z) (17)

Associative symbols: app

Commutative symbols: max', app

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 AC Rule Removal

Using the non-linear polynomial interpretation over the naturals
[max'(x1, x2)] = 1 · x2 + 1 · x1
[app(x1, x2)] = 1 · x2 + 1 · x1 + 2 · x1 · x2
[1] = 0
[s(x1)] = 1 · x1
[0] = 0
[2] = 0
[3] = 0
[4] = 0
[5] = 0
[6] = 2
[7] = 3
[8] = 3
[9] = 3
[empty] = 0
[max(x1)] = 2 · x1
[singl(x1)] = 2 + 1 · x1 + 3 · x1 · x1
[max2(x1, x2)] = 3 + 1 · x2 + 1 · x1 + 2 · x1 · x2
the rules
6 s(5) (6)
7 s(6) (7)
max(singl(x)) x (13)
max(app(singl(x),Y)) max2(x,Y) (14)
max2(x,empty) x (15)
max2(x,singl(y)) max'(x,y) (16)
max2(x,app(singl(y),Z)) max2(max'(x,y),Z) (17)
can be deleted.

1.1 AC Rule Removal

Using the linear polynomial interpretation over the naturals
[max'(x1, x2)] = 1 · x2 + 1 · x1
[app(x1, x2)] = 1 · x2 + 1 · x1
[1] = 0
[s(x1)] = 1 · x1
[0] = 0
[2] = 0
[3] = 1
[4] = 2
[5] = 3
[8] = 0
[7] = 0
[9] = 3
[empty] = 1
the rules
3 s(2) (3)
4 s(3) (4)
5 s(4) (5)
9 s(8) (9)
app(empty,X) X (12)
can be deleted.

1.1.1 AC Rule Removal

Using the non-linear polynomial interpretation over the naturals
[max'(x1, x2)] = 1 · x2 + 1 · x1
[app(x1, x2)] = 3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
[1] = 1
[s(x1)] = 1 + 1 · x1
[0] = 0
[2] = 3
[8] = 1
[7] = 0
the rules
2 s(1) (2)
max'(s(x),s(y)) s(max'(x,y)) (11)
can be deleted.

1.1.1.1 AC Rule Removal

Using the non-linear polynomial interpretation over the naturals
[max'(x1, x2)] = 2 + 1 · x2 + 1 · x1
[app(x1, x2)] = 1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
[1] = 1
[s(x1)] = 2 · x1 · x1
[0] = 0
[8] = 0
[7] = 0
the rules
1 s(0) (1)
max'(0,x) x (10)
can be deleted.

1.1.1.1.1 AC Rule Removal

Using the non-linear polynomial interpretation over the naturals
[max'(x1, x2)] = 1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
[app(x1, x2)] = 1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
[8] = 3
[s(x1)] = 1 + 3 · x1 · x1
[7] = 0
the rule
8 s(7) (8)
can be deleted.

1.1.1.1.1.1 R is empty

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