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,y) -> if(le(x,y),x,y)
  if(true(),x,y) -> |0|()
  if(false(),x,y) -> s(minus(p(x),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,y) -> if#(le(x,y),x,y)
p3: minus#(x,y) -> le#(x,y)
p4: if#(false(),x,y) -> minus#(p(x),y)
p5: if#(false(),x,y) -> p#(x)

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,y) -> if(le(x,y),x,y)
r7: if(true(),x,y) -> |0|()
r8: if(false(),x,y) -> s(minus(p(x),y))

The estimated dependency graph contains the following SCCs:

  {p2, p4}
  {p1}


-- Reduction pair.

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

p1: if#(false(),x,y) -> minus#(p(x),y)
p2: minus#(x,y) -> if#(le(x,y),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,y) -> if(le(x,y),x,y)
r7: if(true(),x,y) -> |0|()
r8: if(false(),x,y) -> s(minus(p(x),y))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      if#_A(x1,x2,x3) = max{x1 + 1, x2 + 4}
      false_A = 5
      minus#_A(x1,x2) = x1 + 4
      p_A(x1) = max{1, x1 - 1}
      le_A(x1,x2) = x1 + 2
      |0|_A = 0
      s_A(x1) = max{5, x1 + 4}
      true_A = 1
    
    2. max/plus interpretations on natural numbers:
    
      if#_A(x1,x2,x3) = 0
      false_A = 0
      minus#_A(x1,x2) = 1
      p_A(x1) = 0
      le_A(x1,x2) = 1
      |0|_A = 0
      s_A(x1) = 0
      true_A = 0
    

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: 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,y) -> if(le(x,y),x,y)
r7: if(true(),x,y) -> |0|()
r8: if(false(),x,y) -> s(minus(p(x),y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      le#_A(x1,x2) = max{x1, x2}
      s_A(x1) = max{2, x1 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      le#_A(x1,x2) = 0
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p1

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