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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      max#_A(x1) = ((1,0),(1,1)) x1
      N_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (1,1)
      L_A(x1) = ((1,0),(1,1)) x1 + (1,1)

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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      max#_A(x1) = ((1,0),(1,0)) x1
      N_A(x1,x2) = x1 + x2
      L_A(x1) = x1 + (1,1)
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1

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