YES
We show the termination of the TRS R:

  times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
  times(x,|0|()) -> |0|()
  times(x,s(y)) -> plus(times(x,y),x)
  plus(x,|0|()) -> x
  plus(x,s(y)) -> s(plus(x,y))


-- SCC decomposition.

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

p1: times#(x,plus(y,s(z))) -> plus#(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
p2: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|())))
p3: times#(x,plus(y,s(z))) -> plus#(y,times(s(z),|0|()))
p4: times#(x,plus(y,s(z))) -> times#(s(z),|0|())
p5: times#(x,plus(y,s(z))) -> times#(x,s(z))
p6: times#(x,s(y)) -> plus#(times(x,y),x)
p7: times#(x,s(y)) -> times#(x,y)
p8: plus#(x,s(y)) -> plus#(x,y)

and R consists of:

r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
r2: times(x,|0|()) -> |0|()
r3: times(x,s(y)) -> plus(times(x,y),x)
r4: plus(x,|0|()) -> x
r5: plus(x,s(y)) -> s(plus(x,y))

The estimated dependency graph contains the following SCCs:

  {p2, p5, p7}
  {p8}


-- Reduction pair.

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

p1: times#(x,s(y)) -> times#(x,y)
p2: times#(x,plus(y,s(z))) -> times#(x,s(z))
p3: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|())))

and R consists of:

r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
r2: times(x,|0|()) -> |0|()
r3: times(x,s(y)) -> plus(times(x,y),x)
r4: plus(x,|0|()) -> x
r5: plus(x,s(y)) -> s(plus(x,y))

The set of usable rules consists of

  r2, r4, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = max{0, x1 - 21, x2 - 25}
            s_A(x1) = max{5, x1}
            plus_A(x1,x2) = max{92, x1 + 19, x2 + 1}
            times_A(x1,x2) = max{91, x2 - 42}
            |0|_A = 65
            times#!_A(x1,x2) = max{1, x2 - 17}
            s!_A(x1) = max{8, x1 - 15}
            plus!_A(x1,x2) = max{104, x1 + 14, x2 - 5}
            times!_A(x1,x2) = max{x1 - 43, x2 + 10}
            |0|!_A = 103
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = max{0, x1 - 7, x2 - 8}
            s_A(x1) = max{41, x1}
            plus_A(x1,x2) = max{7, x1 + 1, x2}
            times_A(x1,x2) = max{29, x1 - 39, x2 - 11}
            |0|_A = 20
            times#!_A(x1,x2) = max{x1 + 66, x2 + 38}
            s!_A(x1) = 0
            plus!_A(x1,x2) = x2 + 2
            times!_A(x1,x2) = max{87, x2 + 45}
            |0|!_A = 44
        
  
    partial status:
  
      pi(times#) = [2]
      pi(s) = [1]
      pi(plus) = [1, 2]
      pi(times) = [2]
      pi(|0|) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- Reduction pair.

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

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

and R consists of:

r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
r2: times(x,|0|()) -> |0|()
r3: times(x,s(y)) -> plus(times(x,y),x)
r4: plus(x,|0|()) -> x
r5: plus(x,s(y)) -> s(plus(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{11, x1 - 15}
            plus#!_A(x1,x2) = max{1, x2 - 12}
            s!_A(x1) = max{2, x1 - 11}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{12, x1 - 5}
            plus#!_A(x1,x2) = max{2, x1 - 1, x2 - 13}
            s!_A(x1) = max{1, x1 - 112}
        
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.