YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    *#_A(x1,x2) = x2 + 1
    +_A(x1,x2) = max{x1, x2 + 1}

The next rules are strictly ordered:

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    *#_A(x1,x2) = x2 + 2
    +_A(x1,x2) = max{x1 + 1, x2 - 1}

The next rules are strictly ordered:

  p1

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:

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

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:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    *#_A(x1,x2) = x1 + 1
    +_A(x1,x2) = max{x1, x2 + 1}

The next rules are strictly ordered:

  p2

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)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    *#_A(x1,x2) = x1
    +_A(x1,x2) = max{x1 + 1, x2 + 1}

The next rules are strictly ordered:

  p1

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