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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |1| > |0| > times > times# > plus
      
      argument filter:
    
        pi(times#) = [1, 2]
        pi(plus) = [1, 2]
        pi(|1|) = []
        pi(times) = 2
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |1| > |0| > times > plus > times#
      
      argument filter:
    
        pi(times#) = [2]
        pi(plus) = 2
        pi(|1|) = []
        pi(times) = 2
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.