YES

We show the termination of the TRS R:

  g(|0|(),f(x,x)) -> x
  g(x,s(y)) -> g(f(x,y),|0|())
  g(s(x),y) -> g(f(x,y),|0|())
  g(f(x,y),|0|()) -> f(g(x,|0|()),g(y,|0|()))

-- SCC decomposition.

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

p1: g#(x,s(y)) -> g#(f(x,y),|0|())
p2: g#(s(x),y) -> g#(f(x,y),|0|())
p3: g#(f(x,y),|0|()) -> g#(x,|0|())
p4: g#(f(x,y),|0|()) -> g#(y,|0|())

and R consists of:

r1: g(|0|(),f(x,x)) -> x
r2: g(x,s(y)) -> g(f(x,y),|0|())
r3: g(s(x),y) -> g(f(x,y),|0|())
r4: g(f(x,y),|0|()) -> f(g(x,|0|()),g(y,|0|()))

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}


-- Reduction pair.

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

p1: g#(f(x,y),|0|()) -> g#(x,|0|())
p2: g#(f(x,y),|0|()) -> g#(y,|0|())
p3: g#(s(x),y) -> g#(f(x,y),|0|())

and R consists of:

r1: g(|0|(),f(x,x)) -> x
r2: g(x,s(y)) -> g(f(x,y),|0|())
r3: g(s(x),y) -> g(f(x,y),|0|())
r4: g(f(x,y),|0|()) -> f(g(x,|0|()),g(y,|0|()))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,1)
            f_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (2,6)
            |0|_A() = (0,1)
            s_A(x1) = ((1,0),(0,0)) x1 + (4,2)
    
      precedence:
      
        g# = f = |0| = s
      
      partial status:
    
        pi(g#) = []
        pi(f) = [2]
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = (0,0)
            f_A(x1,x2) = x2 + (1,2)
            |0|_A() = (1,1)
            s_A(x1) = (2,0)
    
      precedence:
      
        f = |0| = s > g#
      
      partial status:
    
        pi(g#) = []
        pi(f) = [2]
        pi(|0|) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: g(|0|(),f(x,x)) -> x
r2: g(x,s(y)) -> g(f(x,y),|0|())
r3: g(s(x),y) -> g(f(x,y),|0|())
r4: g(f(x,y),|0|()) -> f(g(x,|0|()),g(y,|0|()))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: g(|0|(),f(x,x)) -> x
r2: g(x,s(y)) -> g(f(x,y),|0|())
r3: g(s(x),y) -> g(f(x,y),|0|())
r4: g(f(x,y),|0|()) -> f(g(x,|0|()),g(y,|0|()))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = x1 + (1,2)
            f_A(x1,x2) = ((1,0),(0,0)) x1 + (2,1)
            |0|_A() = (6,0)
            s_A(x1) = ((1,0),(0,0)) x1 + (4,4)
    
      precedence:
      
        g# = f = |0| > s
      
      partial status:
    
        pi(g#) = [1]
        pi(f) = []
        pi(|0|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = (0,0)
            f_A(x1,x2) = (3,0)
            |0|_A() = (2,1)
            s_A(x1) = (1,0)
    
      precedence:
      
        f = s > g# = |0|
      
      partial status:
    
        pi(g#) = []
        pi(f) = []
        pi(|0|) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: g#(s(x),y) -> g#(f(x,y),|0|())

and R consists of:

r1: g(|0|(),f(x,x)) -> x
r2: g(x,s(y)) -> g(f(x,y),|0|())
r3: g(s(x),y) -> g(f(x,y),|0|())
r4: g(f(x,y),|0|()) -> f(g(x,|0|()),g(y,|0|()))

The estimated dependency graph contains the following SCCs:

  (no SCCs)