YES

We show the termination of the TRS R:

  f(|0|()) -> |1|()
  f(s(x)) -> g(x,s(x))
  g(|0|(),y) -> y
  g(s(x),y) -> g(x,+(y,s(x)))
  +(x,|0|()) -> x
  +(x,s(y)) -> s(+(x,y))
  g(s(x),y) -> g(x,s(+(y,x)))

-- SCC decomposition.

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

p1: f#(s(x)) -> g#(x,s(x))
p2: g#(s(x),y) -> g#(x,+(y,s(x)))
p3: g#(s(x),y) -> +#(y,s(x))
p4: +#(x,s(y)) -> +#(x,y)
p5: g#(s(x),y) -> g#(x,s(+(y,x)))
p6: g#(s(x),y) -> +#(y,x)

and R consists of:

r1: f(|0|()) -> |1|()
r2: f(s(x)) -> g(x,s(x))
r3: g(|0|(),y) -> y
r4: g(s(x),y) -> g(x,+(y,s(x)))
r5: +(x,|0|()) -> x
r6: +(x,s(y)) -> s(+(x,y))
r7: g(s(x),y) -> g(x,s(+(y,x)))

The estimated dependency graph contains the following SCCs:

  {p2, p5}
  {p4}


-- Reduction pair.

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

p1: g#(s(x),y) -> g#(x,+(y,s(x)))
p2: g#(s(x),y) -> g#(x,s(+(y,x)))

and R consists of:

r1: f(|0|()) -> |1|()
r2: f(s(x)) -> g(x,s(x))
r3: g(|0|(),y) -> y
r4: g(s(x),y) -> g(x,+(y,s(x)))
r5: +(x,|0|()) -> x
r6: +(x,s(y)) -> s(+(x,y))
r7: g(s(x),y) -> g(x,s(+(y,x)))

The set of usable rules consists of

  r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = x1 + 8
          s_A(x1) = max{1, x1}
          +_A(x1,x2) = max{x1 + 3, x2 + 2}
          |0|_A = 0
    
      precedence:
      
        g# = + > |0| > s
      
      partial status:
    
        pi(g#) = [1]
        pi(s) = [1]
        pi(+) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = max{0, x1 - 3}
          s_A(x1) = x1 + 7
          +_A(x1,x2) = x2 + 5
          |0|_A = 0
    
      precedence:
      
        + > g# = s > |0|
      
      partial status:
    
        pi(g#) = []
        pi(s) = []
        pi(+) = []
        pi(|0|) = []
    

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#(x,s(+(y,x)))

and R consists of:

r1: f(|0|()) -> |1|()
r2: f(s(x)) -> g(x,s(x))
r3: g(|0|(),y) -> y
r4: g(s(x),y) -> g(x,+(y,s(x)))
r5: +(x,|0|()) -> x
r6: +(x,s(y)) -> s(+(x,y))
r7: g(s(x),y) -> g(x,s(+(y,x)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: g#(s(x),y) -> g#(x,s(+(y,x)))

and R consists of:

r1: f(|0|()) -> |1|()
r2: f(s(x)) -> g(x,s(x))
r3: g(|0|(),y) -> y
r4: g(s(x),y) -> g(x,+(y,s(x)))
r5: +(x,|0|()) -> x
r6: +(x,s(y)) -> s(+(x,y))
r7: g(s(x),y) -> g(x,s(+(y,x)))

The set of usable rules consists of

  r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = max{x1 + 6, x2 + 2}
          s_A(x1) = max{3, x1}
          +_A(x1,x2) = max{7, x1, x2 + 4}
          |0|_A = 0
    
      precedence:
      
        g# = + > s = |0|
      
      partial status:
    
        pi(g#) = [1, 2]
        pi(s) = [1]
        pi(+) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = x2 + 1
          s_A(x1) = max{6, x1 + 1}
          +_A(x1,x2) = max{4, x1 + 2, x2 - 2}
          |0|_A = 0
    
      precedence:
      
        g# = s = + = |0|
      
      partial status:
    
        pi(g#) = []
        pi(s) = [1]
        pi(+) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: +#(x,s(y)) -> +#(x,y)

and R consists of:

r1: f(|0|()) -> |1|()
r2: f(s(x)) -> g(x,s(x))
r3: g(|0|(),y) -> y
r4: g(s(x),y) -> g(x,+(y,s(x)))
r5: +(x,|0|()) -> x
r6: +(x,s(y)) -> s(+(x,y))
r7: g(s(x),y) -> g(x,s(+(y,x)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{2, x1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        +# = s
      
      partial status:
    
        pi(+#) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{x1, x2 - 1}
          s_A(x1) = x1
    
      precedence:
      
        +# = s
      
      partial status:
    
        pi(+#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.