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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = max{x1 + 14, x2 + 9}
          false_A = 6
          minus#_A(x1,x2) = max{20, x1 + 10}
          p_A(x1) = max{8, x1 - 2}
          le_A(x1,x2) = max{0, x1 - 5}
          |0|_A = 7
          s_A(x1) = max{13, x1 + 12}
          true_A = 1
    
      precedence:
      
        p > le > if# = false = |0| = true > minus# = s
      
      partial status:
    
        pi(if#) = [1]
        pi(false) = []
        pi(minus#) = [1]
        pi(p) = []
        pi(le) = []
        pi(|0|) = []
        pi(s) = [1]
        pi(true) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = max{3, x1 - 1}
          false_A = 5
          minus#_A(x1,x2) = max{3, x1 + 1}
          p_A(x1) = 1
          le_A(x1,x2) = 0
          |0|_A = 0
          s_A(x1) = x1 + 6
          true_A = 1
    
      precedence:
      
        if# = false = minus# = p = s = true > le = |0|
      
      partial status:
    
        pi(if#) = []
        pi(false) = []
        pi(minus#) = [1]
        pi(p) = []
        pi(le) = []
        pi(|0|) = []
        pi(s) = [1]
        pi(true) = []
    

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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          le#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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