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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              times#_A(x1,x2) = 0
              plus_A(x1,x2) = x1 + 1
              |1|_A() = 1
              times_A(x1,x2) = x1 + 1
              |0|_A() = 1
              times#!_A(x1,x2) = x1 + 1
              plus!_A(x1,x2) = 2
              |1|!_A() = 4
              times!_A(x1,x2) = 3
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = max{0, x1 - 96, x2 - 58}
            plus_A(x1,x2) = max{57, x1 + 10, x2 - 19}
            |1|_A = 52
            times_A(x1,x2) = max{29, x1 - 20, x2 - 24}
            |0|_A = 6
            times#!_A(x1,x2) = max{25, x1 - 51, x2 - 60}
            plus!_A(x1,x2) = max{2, x2 - 31}
            |1|!_A = 26
            times!_A(x1,x2) = max{22, x1 - 29, x2 - 34}
            |0|!_A = 24
        
  
    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.