YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: times#(x,plus(y,s(z))) -> plus#(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
p2: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|())))
p3: times#(x,plus(y,s(z))) -> plus#(y,times(s(z),|0|()))
p4: times#(x,plus(y,s(z))) -> times#(s(z),|0|())
p5: times#(x,plus(y,s(z))) -> times#(x,s(z))
p6: times#(x,s(y)) -> plus#(times(x,y),x)
p7: times#(x,s(y)) -> times#(x,y)
p8: plus#(x,s(y)) -> plus#(x,y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p5, p7}
  {p8}


-- Reduction pair.

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

p1: times#(x,s(y)) -> times#(x,y)
p2: times#(x,plus(y,s(z))) -> times#(x,s(z))
p3: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|())))

and R consists of:

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

The set of usable rules consists of

  r2, r4, r5

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      times#_A(x1,x2) = ((0,1),(0,1)) x2
      s_A(x1) = ((0,1),(0,1)) x1 + (1,1)
      plus_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
      times_A(x1,x2) = ((1,1),(0,0)) x1 + (1,1)
      |0|_A() = (1,1)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- SCC decomposition.

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

p1: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|())))

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  r2, r4, r5

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      times#_A(x1,x2) = ((0,1),(1,1)) x2
      plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,1)) x2
      s_A(x1) = x1 + (1,1)
      times_A(x1,x2) = x1 + ((1,1),(1,1)) x2
      |0|_A() = (0,0)

The next rules are strictly ordered:

  p1

We remove them from the problem.  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)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      plus#_A(x1,x2) = ((1,0),(1,0)) x2
      s_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1

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