YES

We show the termination of the TRS R:

  *(x,*(y,z)) -> *(otimes(x,y),z)
  *(|1|(),y) -> y
  *(+(x,y),z) -> oplus(*(x,z),*(y,z))
  *(x,oplus(y,z)) -> oplus(*(x,y),*(x,z))

-- SCC decomposition.

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

p1: *#(x,*(y,z)) -> *#(otimes(x,y),z)
p2: *#(+(x,y),z) -> *#(x,z)
p3: *#(+(x,y),z) -> *#(y,z)
p4: *#(x,oplus(y,z)) -> *#(x,y)
p5: *#(x,oplus(y,z)) -> *#(x,z)

and R consists of:

r1: *(x,*(y,z)) -> *(otimes(x,y),z)
r2: *(|1|(),y) -> y
r3: *(+(x,y),z) -> oplus(*(x,z),*(y,z))
r4: *(x,oplus(y,z)) -> oplus(*(x,y),*(x,z))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: *#(x,*(y,z)) -> *#(otimes(x,y),z)
p2: *#(x,oplus(y,z)) -> *#(x,z)
p3: *#(x,oplus(y,z)) -> *#(x,y)
p4: *#(+(x,y),z) -> *#(y,z)
p5: *#(+(x,y),z) -> *#(x,z)

and R consists of:

r1: *(x,*(y,z)) -> *(otimes(x,y),z)
r2: *(|1|(),y) -> y
r3: *(+(x,y),z) -> oplus(*(x,z),*(y,z))
r4: *(x,oplus(y,z)) -> oplus(*(x,y),*(x,z))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      *#_A(x1,x2) = x1 + x2
      *_A(x1,x2) = x1 + x2 + 2
      otimes_A(x1,x2) = x1 + x2 + 1
      oplus_A(x1,x2) = x1 + x2
      +_A(x1,x2) = x1 + x2

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

We remove them from the problem.

-- SCC decomposition.

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

p1: *#(x,oplus(y,z)) -> *#(x,z)
p2: *#(x,oplus(y,z)) -> *#(x,y)
p3: *#(+(x,y),z) -> *#(y,z)
p4: *#(+(x,y),z) -> *#(x,z)

and R consists of:

  (no rules)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: *#(x,oplus(y,z)) -> *#(x,z)
p2: *#(+(x,y),z) -> *#(x,z)
p3: *#(+(x,y),z) -> *#(y,z)
p4: *#(x,oplus(y,z)) -> *#(x,y)

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      *#_A(x1,x2) = x2
      oplus_A(x1,x2) = x1 + x2 + 1
      +_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p1, p4

We remove them from the problem.

-- SCC decomposition.

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

p1: *#(+(x,y),z) -> *#(x,z)
p2: *#(+(x,y),z) -> *#(y,z)

and R consists of:

  (no rules)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: *#(+(x,y),z) -> *#(x,z)
p2: *#(+(x,y),z) -> *#(y,z)

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      *#_A(x1,x2) = x1
      +_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p1, p2

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