YES
We show the termination of the TRS R:

  f(x,y,z) -> g(x,y,z)
  g(|0|(),|1|(),x) -> f(x,x,x)
  a() -> b()
  a() -> c()


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(x,y,z) -> g#(x,y,z)
p2: g#(|0|(),|1|(),x) -> f#(x,x,x)

and R consists of:

r1: f(x,y,z) -> g(x,y,z)
r2: g(|0|(),|1|(),x) -> f(x,x,x)
r3: a() -> b()
r4: a() -> c()

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(x,y,z) -> g#(x,y,z)
p2: g#(|0|(),|1|(),x) -> f#(x,x,x)

and R consists of:

r1: f(x,y,z) -> g(x,y,z)
r2: g(|0|(),|1|(),x) -> f(x,x,x)
r3: a() -> b()
r4: a() -> c()

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            f#_A(0,0,0) = 0
            f#_A(0,0,1) = 0
            f#_A(0,1,0) = 0
            f#_A(0,1,1) = 0
            f#_A(1,0,0) = 0
            f#_A(1,0,1) = 0
            f#_A(1,1,0) = 0
            f#_A(1,1,1) = 0
            g#_A(0,0,0) = 0
            g#_A(0,0,1) = 0
            g#_A(0,1,0) = 0
            g#_A(0,1,1) = 0
            g#_A(1,0,0) = 0
            g#_A(1,0,1) = 0
            g#_A(1,1,0) = 0
            g#_A(1,1,1) = 0
            |0|_A = 1
            |1|_A = 0
            f_A(0,0,0) = 0
            f_A(0,0,1) = 0
            f_A(0,1,0) = 0
            f_A(0,1,1) = 0
            f_A(1,0,0) = 0
            f_A(1,0,1) = 0
            f_A(1,1,0) = 0
            f_A(1,1,1) = 0
            g_A(0,0,0) = 0
            g_A(0,0,1) = 0
            g_A(0,1,0) = 0
            g_A(0,1,1) = 0
            g_A(1,0,0) = 0
            g_A(1,0,1) = 0
            g_A(1,1,0) = 0
            g_A(1,1,1) = 0
            a_A = 0
            b_A = 0
            c_A = 0
            f#!_A(0,0,0) = 0
            f#!_A(0,0,1) = 0
            f#!_A(0,1,0) = 0
            f#!_A(0,1,1) = 1
            f#!_A(1,0,0) = 1
            f#!_A(1,0,1) = 1
            f#!_A(1,1,0) = 0
            f#!_A(1,1,1) = 0
            g#!_A(0,0,0) = 0
            g#!_A(0,0,1) = 0
            g#!_A(0,1,0) = 0
            g#!_A(0,1,1) = 1
            g#!_A(1,0,0) = 1
            g#!_A(1,0,1) = 1
            g#!_A(1,1,0) = 0
            g#!_A(1,1,1) = 0
            |0|!_A = 0
            |1|!_A = 1
            f!_A(0,0,0) = 0
            f!_A(0,0,1) = 0
            f!_A(0,1,0) = 0
            f!_A(0,1,1) = 1
            f!_A(1,0,0) = 0
            f!_A(1,0,1) = 0
            f!_A(1,1,0) = 0
            f!_A(1,1,1) = 0
            g!_A(0,0,0) = 0
            g!_A(0,0,1) = 0
            g!_A(0,1,0) = 0
            g!_A(0,1,1) = 1
            g!_A(1,0,0) = 0
            g!_A(1,0,1) = 0
            g!_A(1,1,0) = 0
            g!_A(1,1,1) = 0
            a!_A = 0
            b!_A = 0
            c!_A = 0
  
    partial status:
  
      pi(f#) = []
      pi(g#) = []
      pi(|0|) = []
      pi(|1|) = []
      pi(f) = []
      pi(g) = []
      pi(a) = []
      pi(b) = []
      pi(c) = []

The following rules are removed:

  p2



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(x,y,z) -> g#(x,y,z)

and R consists of:

r1: f(x,y,z) -> g(x,y,z)
r2: g(|0|(),|1|(),x) -> f(x,x,x)
r3: a() -> b()
r4: a() -> c()

The estimated dependency graph contains the following SCCs:

  (no SCCs)