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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        *# > s > p
      
      argument filter:
    
        pi(*#) = [1, 2]
        pi(p) = 1
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        *# > s > p
      
      argument filter:
    
        pi(*#) = [1, 2]
        pi(p) = 1
        pi(s) = 1
    
    3. 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: +(|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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        +# > p > s
      
      argument filter:
    
        pi(+#) = [1, 2]
        pi(s) = 1
        pi(p) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        +# > p > s
      
      argument filter:
    
        pi(+#) = [1, 2]
        pi(s) = 1
        pi(p) = 1
    
    3. 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: +(|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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        minus# > p > s
      
      argument filter:
    
        pi(minus#) = 1
        pi(s) = [1]
        pi(p) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        minus# > p > s
      
      argument filter:
    
        pi(minus#) = 1
        pi(s) = [1]
        pi(p) = 1
    
    3. 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

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