YES

We show the termination of the TRS R:

  p(|0|()) -> |0|()
  p(s(x)) -> x
  le(|0|(),y) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  minus(x,|0|()) -> x
  minus(x,s(y)) -> if(le(x,s(y)),|0|(),p(minus(x,p(s(y)))))
  if(true(),x,y) -> x
  if(false(),x,y) -> y

-- SCC decomposition.

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

p1: le#(s(x),s(y)) -> le#(x,y)
p2: minus#(x,s(y)) -> if#(le(x,s(y)),|0|(),p(minus(x,p(s(y)))))
p3: minus#(x,s(y)) -> le#(x,s(y))
p4: minus#(x,s(y)) -> p#(minus(x,p(s(y))))
p5: minus#(x,s(y)) -> minus#(x,p(s(y)))
p6: minus#(x,s(y)) -> p#(s(y))

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: le(|0|(),y) -> true()
r4: le(s(x),|0|()) -> false()
r5: le(s(x),s(y)) -> le(x,y)
r6: minus(x,|0|()) -> x
r7: minus(x,s(y)) -> if(le(x,s(y)),|0|(),p(minus(x,p(s(y)))))
r8: if(true(),x,y) -> x
r9: if(false(),x,y) -> y

The estimated dependency graph contains the following SCCs:

  {p5}
  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: le(|0|(),y) -> true()
r4: le(s(x),|0|()) -> false()
r5: le(s(x),s(y)) -> le(x,y)
r6: minus(x,|0|()) -> x
r7: minus(x,s(y)) -> if(le(x,s(y)),|0|(),p(minus(x,p(s(y)))))
r8: if(true(),x,y) -> x
r9: if(false(),x,y) -> y

The set of usable rules consists of

  r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        minus#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
        p_A(x1) = x1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        minus#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
        p_A(x1) = 0
    

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

and R consists of:

r1: p(|0|()) -> |0|()
r2: p(s(x)) -> x
r3: le(|0|(),y) -> true()
r4: le(s(x),|0|()) -> false()
r5: le(s(x),s(y)) -> le(x,y)
r6: minus(x,|0|()) -> x
r7: minus(x,s(y)) -> if(le(x,s(y)),|0|(),p(minus(x,p(s(y)))))
r8: if(true(),x,y) -> x
r9: if(false(),x,y) -> y

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        le#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        le#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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