YES

We show the termination of the TRS R:

  max(L(x)) -> x
  max(N(L(|0|()),L(y))) -> y
  max(N(L(s(x)),L(s(y)))) -> s(max(N(L(x),L(y))))
  max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z)))))

-- SCC decomposition.

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

p1: max#(N(L(s(x)),L(s(y)))) -> max#(N(L(x),L(y)))
p2: max#(N(L(x),N(y,z))) -> max#(N(L(x),L(max(N(y,z)))))
p3: max#(N(L(x),N(y,z))) -> max#(N(y,z))

and R consists of:

r1: max(L(x)) -> x
r2: max(N(L(|0|()),L(y))) -> y
r3: max(N(L(s(x)),L(s(y)))) -> s(max(N(L(x),L(y))))
r4: max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z)))))

The estimated dependency graph contains the following SCCs:

  {p3}
  {p1}


-- Reduction pair.

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

p1: max#(N(L(x),N(y,z))) -> max#(N(y,z))

and R consists of:

r1: max(L(x)) -> x
r2: max(N(L(|0|()),L(y))) -> y
r3: max(N(L(s(x)),L(s(y)))) -> s(max(N(L(x),L(y))))
r4: max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z)))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        max# > N > L
      
      argument filter:
    
        pi(max#) = [1]
        pi(N) = [2]
        pi(L) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        N > max# > L
      
      argument filter:
    
        pi(max#) = 1
        pi(N) = 2
        pi(L) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        N > max# > L
      
      argument filter:
    
        pi(max#) = [1]
        pi(N) = 2
        pi(L) = []
    

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: max#(N(L(s(x)),L(s(y)))) -> max#(N(L(x),L(y)))

and R consists of:

r1: max(L(x)) -> x
r2: max(N(L(|0|()),L(y))) -> y
r3: max(N(L(s(x)),L(s(y)))) -> s(max(N(L(x),L(y))))
r4: max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z)))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        L > N > max# > s
      
      argument filter:
    
        pi(max#) = [1]
        pi(N) = [1, 2]
        pi(L) = 1
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        N > max# > s > L
      
      argument filter:
    
        pi(max#) = [1]
        pi(N) = [1, 2]
        pi(L) = []
        pi(s) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        L > N > max# > s
      
      argument filter:
    
        pi(max#) = [1]
        pi(N) = [1, 2]
        pi(L) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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