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:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{11, x2}
        plus_A(x1,x2) = max{11, x1 + 9, x2 + 2}
        |1|_A = 5
        times_A(x1,x2) = max{4, x1 - 8, x2}
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        |1| > plus = times > times# > |0|
      
  
    partial status:
  
      pi(times#) = [2]
      pi(plus) = [2]
      pi(|1|) = []
      pi(times) = [2]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.