YES
We show the termination of the TRS R:

  times(x,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
  times(x,|1|()) -> x
  plus(x,|0|()) -> x
  times(x,|0|()) -> |0|()


-- SCC decomposition.

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

p1: times#(x,plus(y,|1|())) -> plus#(times(x,plus(y,times(|1|(),|0|()))),x)
p2: times#(x,plus(y,|1|())) -> times#(x,plus(y,times(|1|(),|0|())))
p3: times#(x,plus(y,|1|())) -> plus#(y,times(|1|(),|0|()))
p4: times#(x,plus(y,|1|())) -> times#(|1|(),|0|())

and R consists of:

r1: times(x,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
r2: times(x,|1|()) -> x
r3: plus(x,|0|()) -> x
r4: times(x,|0|()) -> |0|()

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: times#(x,plus(y,|1|())) -> times#(x,plus(y,times(|1|(),|0|())))

and R consists of:

r1: times(x,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
r2: times(x,|1|()) -> x
r3: plus(x,|0|()) -> x
r4: times(x,|0|()) -> |0|()

The set of usable rules consists of

  r3, r4

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 - 106, x2 - 42}
            plus_A(x1,x2) = max{x1 + 22, x2 - 17}
            |1|_A = 58
            times_A(x1,x2) = max{10, x1 - 18, x2 - 46}
            |0|_A = 0
            times#!_A(x1,x2) = max{1, x1 - 14, x2 - 36}
            plus!_A(x1,x2) = max{5, x1 - 44, x2 - 122}
            |1|!_A = 4
            times!_A(x1,x2) = max{3, x1 - 71, x2 - 65}
            |0|!_A = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = max{0, x1 - 24, x2 - 27}
            plus_A(x1,x2) = max{11, x1 + 10, x2 + 5}
            |1|_A = 1
            times_A(x1,x2) = max{21, x1 + 4, x2 - 22}
            |0|_A = 21
            times#!_A(x1,x2) = max{2, x1 - 27, x2 - 39}
            plus!_A(x1,x2) = max{0, x2 - 25}
            |1|!_A = 2
            times!_A(x1,x2) = max{1, x1 - 35, x2 - 27}
            |0|!_A = 0
        
  
    partial status:
  
      pi(times#) = [2]
      pi(plus) = [1, 2]
      pi(|1|) = []
      pi(times) = [2]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.