YES

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

Consider the left-linear TRS R:

  s(p(x)) -> x
  p(s(x)) -> x
  +(x,0()) -> x
  +(x,s(y)) -> s(+(x,y))
  +(x,p(y)) -> p(+(x,y))
  -(0(),0()) -> 0()
  -(x,s(y)) -> p(-(x,y))
  -(x,p(y)) -> s(-(x,y))
  -(p(x),y) -> p(-(x,y))
  -(s(x),y) -> s(-(x,y))

Let C be the following subset of R:

  s(p(x)) -> x
  p(s(x)) -> x
  +(x,0()) -> x
  +(x,s(y)) -> s(+(x,y))
  +(x,p(y)) -> p(+(x,y))
  -(0(),0()) -> 0()
  -(x,s(y)) -> p(-(x,y))
  -(x,p(y)) -> s(-(x,y))
  -(p(x),y) -> p(-(x,y))
  -(s(x),y) -> s(-(x,y))

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

  phi(s(p(x)) -> x) = 0
  phi(p(s(x)) -> x) = 0
  phi(+(x,0()) -> x) = 0
  phi(+(x,s(y)) -> s(+(x,y))) = 0
  phi(+(x,p(y)) -> p(+(x,y))) = 0
  phi(-(0(),0()) -> 0()) = 0
  phi(-(x,s(y)) -> p(-(x,y))) = 0
  phi(-(x,p(y)) -> s(-(x,y))) = 0
  phi(-(p(x),y) -> p(-(x,y))) = 0
  phi(-(s(x),y) -> s(-(x,y))) = 0

  psi(s(p(x)) -> x) = 0
  psi(p(s(x)) -> x) = 0
  psi(+(x,0()) -> x) = 0
  psi(+(x,s(y)) -> s(+(x,y))) = 0
  psi(+(x,p(y)) -> p(+(x,y))) = 0
  psi(-(0(),0()) -> 0()) = 0
  psi(-(x,s(y)) -> p(-(x,y))) = 0
  psi(-(x,p(y)) -> s(-(x,y))) = 0
  psi(-(p(x),y) -> p(-(x,y))) = 0
  psi(-(s(x),y) -> s(-(x,y))) = 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:

  s(p(x)) -> x
  p(s(x)) -> x
  +(x,0()) -> x
  +(x,s(y)) -> s(+(x,y))
  +(x,p(y)) -> p(+(x,y))
  -(0(),0()) -> 0()
  -(x,s(y)) -> p(-(x,y))
  -(x,p(y)) -> s(-(x,y))
  -(p(x),y) -> p(-(x,y))
  -(s(x),y) -> s(-(x,y))

Let C be the following subset of R:

  (empty)

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

  +(y1,s(p(x1_1))) -> +(y1,x1_1)
  +(y1,s(p(x1_1))) -> s(+(y1,p(x1_1)))
  +(y1,p(s(x1_1))) -> +(y1,x1_1)
  +(y1,p(s(x1_1))) -> p(+(y1,s(x1_1)))
  -(p(x0_1),s(y2)) -> p(-(x0_1,s(y2)))
  -(p(x0_1),s(y2)) -> p(-(p(x0_1),y2))
  -(s(x0_1),s(y2)) -> s(-(x0_1,s(y2)))
  -(s(x0_1),s(y2)) -> p(-(s(x0_1),y2))
  -(y1,s(p(x1_1))) -> -(y1,x1_1)
  -(y1,s(p(x1_1))) -> p(-(y1,p(x1_1)))
  -(p(x0_1),p(y2)) -> p(-(x0_1,p(y2)))
  -(p(x0_1),p(y2)) -> s(-(p(x0_1),y2))
  -(s(x0_1),p(y2)) -> s(-(x0_1,p(y2)))
  -(s(x0_1),p(y2)) -> s(-(s(x0_1),y2))
  -(y1,p(s(x1_1))) -> -(y1,x1_1)
  -(y1,p(s(x1_1))) -> s(-(y1,s(x1_1)))
  -(p(y1),s(x0_2)) -> p(-(p(y1),x0_2))
  -(p(y1),s(x0_2)) -> p(-(y1,s(x0_2)))
  -(p(y1),p(x0_2)) -> s(-(p(y1),x0_2))
  -(p(y1),p(x0_2)) -> p(-(y1,p(x0_2)))
  -(p(s(x1_1)),y2) -> -(x1_1,y2)
  -(p(s(x1_1)),y2) -> p(-(s(x1_1),y2))
  -(s(y1),s(x0_2)) -> p(-(s(y1),x0_2))
  -(s(y1),s(x0_2)) -> s(-(y1,s(x0_2)))
  -(s(y1),p(x0_2)) -> s(-(s(y1),x0_2))
  -(s(y1),p(x0_2)) -> s(-(y1,p(x0_2)))
  -(s(p(x1_1)),y2) -> -(x1_1,y2)
  -(s(p(x1_1)),y2) -> s(-(p(x1_1),y2))

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.