YES

# parallel critical pair closing system (Shintani and Hirokawa 2022)

Consider the left-linear TRS R:

  a1() -> b1()
  a1() -> c1()
  b1() -> b2()
  c1() -> c2()
  a2() -> b2()
  a2() -> c2()
  b2() -> b3()
  c2() -> c3()
  a3() -> b3()
  a3() -> c3()
  b3() -> b4()
  c3() -> c4()
  a4() -> b4()
  a4() -> c4()
  b4() -> b5()
  c4() -> c5()
  a5() -> b6()
  b5() -> b6()
  c5() -> b6()

Let C be the following subset of R:

  b1() -> b2()
  c1() -> c2()
  b2() -> b3()
  c2() -> c3()
  b3() -> b4()
  c3() -> c4()
  b4() -> b5()
  c4() -> c5()
  b5() -> b6()
  c5() -> b6()

The TRS R is left-linear and all parallel critical pairs are joinable by C.
Therefore, the confluence of R follows from that of C.

# parallel critical pair closing system (Shintani and Hirokawa 2022)

Consider the left-linear TRS R:

  b1() -> b2()
  c1() -> c2()
  b2() -> b3()
  c2() -> c3()
  b3() -> b4()
  c3() -> c4()
  b4() -> b5()
  c4() -> c5()
  b5() -> b6()
  c5() -> b6()

Let C be the following subset of R:

  (empty)

The TRS R is left-linear and all parallel critical pairs are joinable by C.
Therefore, the confluence of R follows from that of C.

# emptiness

The empty TRS is confluent.