Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z078)

The rewrite relation of the following TRS is considered.

f(0(x1)) → s(0(x1)) (1)
d(0(x1)) → 0(x1) (2)
d(s(x1)) → s(s(d(p(s(x1))))) (3)
f(s(x1)) → d(f(p(s(x1)))) (4)
p(s(x1)) → x1 (5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
0(f(x1)) → 0(s(x1)) (6)
0(d(x1)) → 0(x1) (7)
s(d(x1)) → s(p(d(s(s(x1))))) (8)
s(f(x1)) → s(p(f(d(x1)))) (9)
s(p(x1)) → x1 (10)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[p(x1)] = 0 · x1 + -∞
[f(x1)] = 13 · x1 + -∞
[d(x1)] = 0 · x1 + -∞
[0(x1)] = 4 · x1 + -∞
[s(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
0(f(x1)) → 0(s(x1)) (6)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[p(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
0 1 0
0 1 0
· x1 +
0 0 0
0 0 0
1 0 0
[d(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[0(x1)] =
1 0 1
0 0 1
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
0(d(x1)) → 0(x1) (7)

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
s#(d(x1)) → s#(x1) (11)
s#(d(x1)) → s#(s(x1)) (12)
s#(d(x1)) → s#(p(d(s(s(x1))))) (13)
s#(f(x1)) → s#(p(f(d(x1)))) (14)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.