YES

We show the termination of the TRS R:

  +(|0|(),y) -> y
  +(s(x),y) -> s(+(x,y))
  +(p(x),y) -> p(+(x,y))
  minus(|0|()) -> |0|()
  minus(s(x)) -> p(minus(x))
  minus(p(x)) -> s(minus(x))
  *(|0|(),y) -> |0|()
  *(s(x),y) -> +(*(x,y),y)
  *(p(x),y) -> +(*(x,y),minus(y))

-- SCC decomposition.

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

p1: +#(s(x),y) -> +#(x,y)
p2: +#(p(x),y) -> +#(x,y)
p3: minus#(s(x)) -> minus#(x)
p4: minus#(p(x)) -> minus#(x)
p5: *#(s(x),y) -> +#(*(x,y),y)
p6: *#(s(x),y) -> *#(x,y)
p7: *#(p(x),y) -> +#(*(x,y),minus(y))
p8: *#(p(x),y) -> *#(x,y)
p9: *#(p(x),y) -> minus#(y)

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: +(p(x),y) -> p(+(x,y))
r4: minus(|0|()) -> |0|()
r5: minus(s(x)) -> p(minus(x))
r6: minus(p(x)) -> s(minus(x))
r7: *(|0|(),y) -> |0|()
r8: *(s(x),y) -> +(*(x,y),y)
r9: *(p(x),y) -> +(*(x,y),minus(y))

The estimated dependency graph contains the following SCCs:

  {p6, p8}
  {p1, p2}
  {p3, p4}


-- Reduction pair.

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

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

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: +(p(x),y) -> p(+(x,y))
r4: minus(|0|()) -> |0|()
r5: minus(s(x)) -> p(minus(x))
r6: minus(p(x)) -> s(minus(x))
r7: *(|0|(),y) -> |0|()
r8: *(s(x),y) -> +(*(x,y),y)
r9: *(p(x),y) -> +(*(x,y),minus(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2

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: +#(s(x),y) -> +#(x,y)
p2: +#(p(x),y) -> +#(x,y)

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: +(p(x),y) -> p(+(x,y))
r4: minus(|0|()) -> |0|()
r5: minus(s(x)) -> p(minus(x))
r6: minus(p(x)) -> s(minus(x))
r7: *(|0|(),y) -> |0|()
r8: *(s(x),y) -> +(*(x,y),y)
r9: *(p(x),y) -> +(*(x,y),minus(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2

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: minus#(s(x)) -> minus#(x)
p2: minus#(p(x)) -> minus#(x)

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: +(p(x),y) -> p(+(x,y))
r4: minus(|0|()) -> |0|()
r5: minus(s(x)) -> p(minus(x))
r6: minus(p(x)) -> s(minus(x))
r7: *(|0|(),y) -> |0|()
r8: *(s(x),y) -> +(*(x,y),y)
r9: *(p(x),y) -> +(*(x,y),minus(y))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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