YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

p1: times#(x,s(y)) -> plus#(times(x,y),x)
p2: times#(x,s(y)) -> times#(x,y)
p3: plus#(x,s(y)) -> plus#(x,y)
p4: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(x,|0|()) -> x
r4: plus(|0|(),x) -> x
r5: plus(x,s(y)) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(x,y))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3, p4}


-- Reduction pair.

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

p1: times#(x,s(y)) -> times#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(x,|0|()) -> x
r4: plus(|0|(),x) -> x
r5: plus(x,s(y)) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{4, x2 + 2}
        s_A(x1) = max{14, x1 + 3}
  
    base order 2:
  
      precedence:
      
        s > times#
      
  
    partial status:
  
      pi(times#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: plus#(x,s(y)) -> plus#(x,y)
p2: plus#(s(x),y) -> plus#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(x,|0|()) -> x
r4: plus(|0|(),x) -> x
r5: plus(x,s(y)) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(x,y))

The set of usable rules consists of

  r3

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x1 + 2
        s_A(x1) = max{5, x1 + 3}
        plus_A(x1,x2) = max{1, x1}
        |0|_A = 3
  
    base order 2:
  
      precedence:
      
        s > plus# = |0| > plus
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []
      pi(plus) = [1]
      pi(|0|) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: plus#(x,s(y)) -> plus#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(x,|0|()) -> x
r4: plus(|0|(),x) -> x
r5: plus(x,s(y)) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(x,y))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: plus#(x,s(y)) -> plus#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(x,|0|()) -> x
r4: plus(|0|(),x) -> x
r5: plus(x,s(y)) -> s(plus(x,y))
r6: plus(s(x),y) -> s(plus(x,y))

The set of usable rules consists of

  r3

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{x1 + 4, x2 + 4}
        s_A(x1) = max{2, x1}
        plus_A(x1,x2) = max{4, x1 + 1, x2 - 3}
        |0|_A = 0
  
    base order 2:
  
      precedence:
      
        plus > plus# > s = |0|
      
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = [1]
      pi(plus) = [1]
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.