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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        max#_A(x1) = max{5, x1 + 3}
        N_A(x1,x2) = max{x1 + 2, x2 + 2}
        L_A(x1) = x1 + 2
  
    precedence:
    
      max# = N = L
    
    partial status:
  
      pi(max#) = []
      pi(N) = [2]
      pi(L) = [1]

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        max#_A(x1) = x1 + 12
        N_A(x1,x2) = max{x1 - 6, x2 + 1}
        L_A(x1) = x1 + 4
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      max# = N = L = s
    
    partial status:
  
      pi(max#) = []
      pi(N) = [2]
      pi(L) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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