YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: exp#(x,s(y)) -> *#(x,exp(x,y))
p2: exp#(x,s(y)) -> exp#(x,y)
p3: *#(s(x),y) -> *#(x,y)
p4: -#(s(x),s(y)) -> -#(x,y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3}
  {p4}


-- Reduction pair.

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

p1: exp#(x,s(y)) -> exp#(x,y)

and R consists of:

r1: exp(x,|0|()) -> s(|0|())
r2: exp(x,s(y)) -> *(x,exp(x,y))
r3: *(|0|(),y) -> |0|()
r4: *(s(x),y) -> +(y,*(x,y))
r5: -(|0|(),y) -> |0|()
r6: -(x,|0|()) -> x
r7: -(s(x),s(y)) -> -(x,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:
      
        exp# > s
      
      argument filter:
    
        pi(exp#) = 2
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > exp#
      
      argument filter:
    
        pi(exp#) = 2
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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)

and R consists of:

r1: exp(x,|0|()) -> s(|0|())
r2: exp(x,s(y)) -> *(x,exp(x,y))
r3: *(|0|(),y) -> |0|()
r4: *(s(x),y) -> +(y,*(x,y))
r5: -(|0|(),y) -> |0|()
r6: -(x,|0|()) -> x
r7: -(s(x),s(y)) -> -(x,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 > *#
      
      argument filter:
    
        pi(*#) = 1
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > *#
      
      argument filter:
    
        pi(*#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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),s(y)) -> -#(x,y)

and R consists of:

r1: exp(x,|0|()) -> s(|0|())
r2: exp(x,s(y)) -> *(x,exp(x,y))
r3: *(|0|(),y) -> |0|()
r4: *(s(x),y) -> +(y,*(x,y))
r5: -(|0|(),y) -> |0|()
r6: -(x,|0|()) -> x
r7: -(s(x),s(y)) -> -(x,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 > -#
      
      argument filter:
    
        pi(-#) = [2]
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > -#
      
      argument filter:
    
        pi(-#) = [2]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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