YES

We show the termination of the TRS R:

  p(s(x)) -> x
  s(p(x)) -> x
  +(|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) -> s#(+(x,y))
p2: +#(s(x),y) -> +#(x,y)
p3: +#(p(x),y) -> p#(+(x,y))
p4: +#(p(x),y) -> +#(x,y)
p5: minus#(s(x)) -> p#(minus(x))
p6: minus#(s(x)) -> minus#(x)
p7: minus#(p(x)) -> s#(minus(x))
p8: minus#(p(x)) -> minus#(x)
p9: *#(s(x),y) -> +#(*(x,y),y)
p10: *#(s(x),y) -> *#(x,y)
p11: *#(p(x),y) -> +#(*(x,y),minus(y))
p12: *#(p(x),y) -> *#(x,y)
p13: *#(p(x),y) -> minus#(y)

and R consists of:

r1: p(s(x)) -> x
r2: s(p(x)) -> x
r3: +(|0|(),y) -> y
r4: +(s(x),y) -> s(+(x,y))
r5: +(p(x),y) -> p(+(x,y))
r6: minus(|0|()) -> |0|()
r7: minus(s(x)) -> p(minus(x))
r8: minus(p(x)) -> s(minus(x))
r9: *(|0|(),y) -> |0|()
r10: *(s(x),y) -> +(*(x,y),y)
r11: *(p(x),y) -> +(*(x,y),minus(y))

The estimated dependency graph contains the following SCCs:

  {p10, p12}
  {p2, p4}
  {p6, p8}


-- 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: p(s(x)) -> x
r2: s(p(x)) -> x
r3: +(|0|(),y) -> y
r4: +(s(x),y) -> s(+(x,y))
r5: +(p(x),y) -> p(+(x,y))
r6: minus(|0|()) -> |0|()
r7: minus(s(x)) -> p(minus(x))
r8: minus(p(x)) -> s(minus(x))
r9: *(|0|(),y) -> |0|()
r10: *(s(x),y) -> +(*(x,y),y)
r11: *(p(x),y) -> +(*(x,y),minus(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      *# > s > p
    
    argument filter:
  
      pi(*#) = 1
      pi(p) = [1]
      pi(s) = [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: p(s(x)) -> x
r2: s(p(x)) -> x
r3: +(|0|(),y) -> y
r4: +(s(x),y) -> s(+(x,y))
r5: +(p(x),y) -> p(+(x,y))
r6: minus(|0|()) -> |0|()
r7: minus(s(x)) -> p(minus(x))
r8: minus(p(x)) -> s(minus(x))
r9: *(|0|(),y) -> |0|()
r10: *(s(x),y) -> +(*(x,y),y)
r11: *(p(x),y) -> +(*(x,y),minus(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      +# > p > s
    
    argument filter:
  
      pi(+#) = 1
      pi(s) = [1]
      pi(p) = [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: p(s(x)) -> x
r2: s(p(x)) -> x
r3: +(|0|(),y) -> y
r4: +(s(x),y) -> s(+(x,y))
r5: +(p(x),y) -> p(+(x,y))
r6: minus(|0|()) -> |0|()
r7: minus(s(x)) -> p(minus(x))
r8: minus(p(x)) -> s(minus(x))
r9: *(|0|(),y) -> |0|()
r10: *(s(x),y) -> +(*(x,y),y)
r11: *(p(x),y) -> +(*(x,y),minus(y))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      minus# > p > s
    
    argument filter:
  
      pi(minus#) = 1
      pi(s) = [1]
      pi(p) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

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