YES

# Compositional parallel critical pair system (Shintani and Hirokawa 2022).

Consider the left-linear TRS R:

  foo(0(x)) -> 0(s(p(p(p(s(s(s(p(s(x))))))))))
  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))))))))))))))))))))))))))
  bar(0(x)) -> 0(p(s(s(s(x)))))
  bar(s(x)) -> p(s(p(p(s(s(foo(s(p(p(s(s(x))))))))))))
  p(p(s(x))) -> p(x)
  p(s(x)) -> x
  p(0(x)) -> 0(s(s(s(s(x)))))

Let C be the following subset of R:

  foo(0(x)) -> 0(s(p(p(p(s(s(s(p(s(x))))))))))
  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))))))))))))))))))))))))))
  bar(0(x)) -> 0(p(s(s(s(x)))))
  bar(s(x)) -> p(s(p(p(s(s(foo(s(p(p(s(s(x))))))))))))
  p(p(s(x))) -> p(x)
  p(s(x)) -> x
  p(0(x)) -> 0(s(s(s(s(x)))))

The parallel critical pair system PCPS(R,C) is:

  (empty)

All pairs in PCP(R) are joinable and PCPS(R,C)/R is terminating.
Therefore, the confluence of R follows from that of C.

# Parallel rule labeling (Zankl et al. 2015).

Consider the left-linear TRS R:

  foo(0(x)) -> 0(s(p(p(p(s(s(s(p(s(x))))))))))
  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))))))))))))))))))))))))))
  bar(0(x)) -> 0(p(s(s(s(x)))))
  bar(s(x)) -> p(s(p(p(s(s(foo(s(p(p(s(s(x))))))))))))
  p(p(s(x))) -> p(x)
  p(s(x)) -> x
  p(0(x)) -> 0(s(s(s(s(x)))))

All parallel critical peaks (except C's) are decreasing wrt rule labeling:

  phi(foo(0(x)) -> 0(s(p(p(p(s(s(s(p(s(x))))))))))) = 1
  phi(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))))))))))))))))))))))))))) = 1
  phi(bar(0(x)) -> 0(p(s(s(s(x)))))) = 1
  phi(bar(s(x)) -> p(s(p(p(s(s(foo(s(p(p(s(s(x))))))))))))) = 1
  phi(p(p(s(x))) -> p(x)) = 1
  phi(p(s(x)) -> x) = 1
  phi(p(0(x)) -> 0(s(s(s(s(x)))))) = 1

  psi(foo(0(x)) -> 0(s(p(p(p(s(s(s(p(s(x))))))))))) = 1
  psi(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))))))))))))))))))))))))))) = 1
  psi(bar(0(x)) -> 0(p(s(s(s(x)))))) = 1
  psi(bar(s(x)) -> p(s(p(p(s(s(foo(s(p(p(s(s(x))))))))))))) = 1
  psi(p(p(s(x))) -> p(x)) = 1
  psi(p(s(x)) -> x) = 1
  psi(p(0(x)) -> 0(s(s(s(s(x)))))) = 1