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, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          times#_A(x1,x2) = x1 + x2
          plus_A(x1,x2) = x1 + x2 + 2
          |1|_A() = 4
          times_A(x1,x2) = 3
          |0|_A() = 2
  
    base order 2:
  
      precedence:
      
        times# = plus = |1| = times > |0|
      
  
    partial status:
  
      pi(times#) = [1]
      pi(plus) = [1]
      pi(|1|) = []
      pi(times) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.