YES

# Compositional parallel rule labeling (Shintani and Hirokawa 2022).

Consider the left-linear TRS R:

  a(s(x)) -> s(a(x))
  b(a(b(s(x)))) -> a(b(s(a(x))))
  b(a(b(b(x)))) -> a(b(a(b(x))))
  a(b(a(a(x)))) -> b(a(b(a(x))))

Let C be the following subset of R:

  a(s(x)) -> s(a(x))
  b(a(b(s(x)))) -> a(b(s(a(x))))
  b(a(b(b(x)))) -> a(b(a(b(x))))
  a(b(a(a(x)))) -> b(a(b(a(x))))

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

  phi(a(s(x)) -> s(a(x))) = 0
  phi(b(a(b(s(x)))) -> a(b(s(a(x))))) = 0
  phi(b(a(b(b(x)))) -> a(b(a(b(x))))) = 0
  phi(a(b(a(a(x)))) -> b(a(b(a(x))))) = 0

  psi(a(s(x)) -> s(a(x))) = 0
  psi(b(a(b(s(x)))) -> a(b(s(a(x))))) = 0
  psi(b(a(b(b(x)))) -> a(b(a(b(x))))) = 0
  psi(a(b(a(a(x)))) -> b(a(b(a(x))))) = 0


Therefore, the confluence of R follows from that of C.

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

Consider the left-linear TRS R:

  a(s(x)) -> s(a(x))
  b(a(b(s(x)))) -> a(b(s(a(x))))
  b(a(b(b(x)))) -> a(b(a(b(x))))
  a(b(a(a(x)))) -> b(a(b(a(x))))

Let C be the following subset of R:

  (empty)

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

  b(a(b(b(a(b(s(x3_1))))))) -> b(a(b(a(b(s(a(x3_1)))))))
  b(a(b(b(a(b(s(x3_1))))))) -> a(b(a(b(a(b(s(x3_1)))))))
  b(a(b(b(a(b(b(x3_1))))))) -> b(a(b(a(b(a(b(x3_1)))))))
  b(a(b(b(a(b(b(x3_1))))))) -> a(b(a(b(a(b(b(x3_1)))))))
  a(b(a(a(s(x3_1))))) -> a(b(a(s(a(x3_1)))))
  a(b(a(a(s(x3_1))))) -> b(a(b(a(s(x3_1)))))
  a(b(a(a(b(a(a(x3_1))))))) -> a(b(a(b(a(b(a(x3_1)))))))
  a(b(a(a(b(a(a(x3_1))))))) -> b(a(b(a(b(a(a(x3_1)))))))

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

# emptiness

The empty TRS is confluent.