YES

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

Consider the left-linear TRS R:

  if(true(),x,y) -> x
  if(false(),x,y) -> y
  -(s(x),s(y)) -> -(x,y)
  -(x,0()) -> x
  -(0(),x) -> 0()
  <(s(x),s(y)) -> <(x,y)
  <(0(),s(x)) -> true()
  <(x,0()) -> false()
  mod(0(),y) -> 0()
  mod(x,s(y)) -> if(<(x,s(y)),x,mod(-(x,s(y)),s(y)))
  mod(x,0()) -> x
  gcd(x,y) -> gcd(y,mod(x,y))
  gcd(x,0()) -> x
  gcd(0(),x) -> x

Let C be the following subset of R:

  (empty)

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

  phi(if(true(),x,y) -> x) = 1
  phi(if(false(),x,y) -> y) = 1
  phi(-(s(x),s(y)) -> -(x,y)) = 1
  phi(-(x,0()) -> x) = 1
  phi(-(0(),x) -> 0()) = 1
  phi(<(s(x),s(y)) -> <(x,y)) = 1
  phi(<(0(),s(x)) -> true()) = 1
  phi(<(x,0()) -> false()) = 1
  phi(mod(0(),y) -> 0()) = 4
  phi(mod(x,s(y)) -> if(<(x,s(y)),x,mod(-(x,s(y)),s(y)))) = 4
  phi(mod(x,0()) -> x) = 5
  phi(gcd(x,y) -> gcd(y,mod(x,y))) = 4
  phi(gcd(x,0()) -> x) = 6
  phi(gcd(0(),x) -> x) = 6

  psi(if(true(),x,y) -> x) = 3
  psi(if(false(),x,y) -> y) = 1
  psi(-(s(x),s(y)) -> -(x,y)) = 1
  psi(-(x,0()) -> x) = 1
  psi(-(0(),x) -> 0()) = 1
  psi(<(s(x),s(y)) -> <(x,y)) = 1
  psi(<(0(),s(x)) -> true()) = 2
  psi(<(x,0()) -> false()) = 1
  psi(mod(0(),y) -> 0()) = 2
  psi(mod(x,s(y)) -> if(<(x,s(y)),x,mod(-(x,s(y)),s(y)))) = 2
  psi(mod(x,0()) -> x) = 1
  psi(gcd(x,y) -> gcd(y,mod(x,y))) = 5
  psi(gcd(x,0()) -> x) = 3
  psi(gcd(0(),x) -> x) = 3


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:

  (empty)

Let C be the following subset of R:

  (empty)

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.

# emptiness

The empty TRS is confluent.