Certification Problem

Input (COPS 977)

We consider the TRS containing the following rules:

foo(0(x)) 0(s(p(p(p(s(s(s(p(s(x)))))))))) (1)
foo(s(x)) p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x)))))))))))))))))))))))))) (2)
bar(0(x)) 0(p(s(s(s(x))))) (3)
bar(s(x)) p(s(p(p(s(s(foo(s(p(p(s(s(x)))))))))))) (4)
p(p(s(x))) p(x) (5)
p(s(x)) x (6)
p(0(x)) 0(s(s(s(s(x))))) (7)

The underlying signature is as follows:

{foo/1, 0/1, s/1, p/1, bar/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2021)

1 Critical Pair Closing System

Confluence is proven using the following terminating critical-pair-closing-system R:

There are no rules.

1.1 R is empty

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