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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        if#_A(x1,x2,x3) = max{x1, x2 + 5}
        false_A = 10
        minus#_A(x1,x2) = x1 + 5
        p_A(x1) = max{4, x1 - 1}
        le_A(x1,x2) = max{5, x1 + 4}
        |0|_A = 3
        s_A(x1) = x1 + 11
        true_A = 6
  
    precedence:
    
      p > false = minus# = le > if# > |0| = s = true
    
    partial status:
  
      pi(if#) = []
      pi(false) = []
      pi(minus#) = []
      pi(p) = []
      pi(le) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(true) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: if#(false(),x,y) -> minus#(p(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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        le#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      le# = s
    
    partial status:
  
      pi(le#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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