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 reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      *# > + > oplus > otimes > *
    
    argument filter:
  
      pi(*#) = 2
      pi(*) = [1, 2]
      pi(otimes) = 2
      pi(oplus) = [1, 2]
      pi(+) = [1, 2]

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.

-- SCC decomposition.

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

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

  lexicographic path order with precedence:
  
    precedence:
    
      + > *#
    
    argument filter:
  
      pi(*#) = [1]
      pi(+) = [1, 2]

The next rules are strictly ordered:

  p1, p2

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