YES

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

Consider the left-linear TRS R:

  f(g(x,a(),b())) -> x
  g(f(h(c(),d())),x,y) -> h(k1(x),k2(y))
  k1(a()) -> c()
  k2(b()) -> d()
  f(h(k1(a()),k2(b()))) -> f(h(c(),d()))
  f(h(c(),k2(b()))) -> f(h(c(),d()))
  f(h(k1(a()),d())) -> f(h(c(),d()))

Let C be the following subset of R:

  f(g(x,a(),b())) -> x
  g(f(h(c(),d())),x,y) -> h(k1(x),k2(y))
  k1(a()) -> c()
  k2(b()) -> d()
  f(h(k1(a()),k2(b()))) -> f(h(c(),d()))
  f(h(c(),k2(b()))) -> f(h(c(),d()))
  f(h(k1(a()),d())) -> f(h(c(),d()))

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:

  f(g(x,a(),b())) -> x
  g(f(h(c(),d())),x,y) -> h(k1(x),k2(y))
  k1(a()) -> c()
  k2(b()) -> d()
  f(h(k1(a()),k2(b()))) -> f(h(c(),d()))
  f(h(c(),k2(b()))) -> f(h(c(),d()))
  f(h(k1(a()),d())) -> f(h(c(),d()))

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

  phi(f(g(x,a(),b())) -> x) = 4
  phi(g(f(h(c(),d())),x,y) -> h(k1(x),k2(y))) = 4
  phi(k1(a()) -> c()) = 2
  phi(k2(b()) -> d()) = 2
  phi(f(h(k1(a()),k2(b()))) -> f(h(c(),d()))) = 3
  phi(f(h(c(),k2(b()))) -> f(h(c(),d()))) = 1
  phi(f(h(k1(a()),d())) -> f(h(c(),d()))) = 1

  psi(f(g(x,a(),b())) -> x) = 4
  psi(g(f(h(c(),d())),x,y) -> h(k1(x),k2(y))) = 4
  psi(k1(a()) -> c()) = 4
  psi(k2(b()) -> d()) = 3
  psi(f(h(k1(a()),k2(b()))) -> f(h(c(),d()))) = 5
  psi(f(h(c(),k2(b()))) -> f(h(c(),d()))) = 1
  psi(f(h(k1(a()),d())) -> f(h(c(),d()))) = 1