YES

We show the termination of the TRS R:

  le(|0|(),y) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  minus(|0|(),y) -> |0|()
  minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
  if_minus(true(),s(x),y) -> |0|()
  if_minus(false(),s(x),y) -> s(minus(x,y))
  mod(|0|(),y) -> |0|()
  mod(s(x),|0|()) -> |0|()
  mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
  if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
  if_mod(false(),s(x),s(y)) -> s(x)

-- SCC decomposition.

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

p1: le#(s(x),s(y)) -> le#(x,y)
p2: minus#(s(x),y) -> if_minus#(le(s(x),y),s(x),y)
p3: minus#(s(x),y) -> le#(s(x),y)
p4: if_minus#(false(),s(x),y) -> minus#(x,y)
p5: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))
p6: mod#(s(x),s(y)) -> le#(y,x)
p7: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p8: if_mod#(true(),s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)

The estimated dependency graph contains the following SCCs:

  {p5, p7}
  {p2, p4}
  {p1}


-- Reduction pair.

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

p1: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p2: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_mod#_A(x1,x2,x3) = max{x1 + 20, x2 + 1}
          true_A = 17
          s_A(x1) = x1 + 63
          mod#_A(x1,x2) = max{64, x1 + 5}
          minus_A(x1,x2) = x1 + 59
          le_A(x1,x2) = x2 + 38
          if_minus_A(x1,x2,x3) = x2 + 59
          |0|_A = 0
          false_A = 37
    
      precedence:
      
        if_mod# > true = mod# = minus = le > if_minus > s = |0| = false
      
      partial status:
    
        pi(if_mod#) = [1, 2]
        pi(true) = []
        pi(s) = [1]
        pi(mod#) = [1]
        pi(minus) = [1]
        pi(le) = []
        pi(if_minus) = [2]
        pi(|0|) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_mod#_A(x1,x2,x3) = max{1, x1 - 6, x2 - 5}
          true_A = 16
          s_A(x1) = max{14, x1 + 10}
          mod#_A(x1,x2) = x1
          minus_A(x1,x2) = max{2, x1}
          le_A(x1,x2) = 15
          if_minus_A(x1,x2,x3) = max{15, x2 + 5}
          |0|_A = 3
          false_A = 9
    
      precedence:
      
        if_mod# = true = s = mod# = minus = le = if_minus = |0| = false
      
      partial status:
    
        pi(if_mod#) = []
        pi(true) = []
        pi(s) = [1]
        pi(mod#) = [1]
        pi(minus) = [1]
        pi(le) = []
        pi(if_minus) = []
        pi(|0|) = []
        pi(false) = []
    

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: if_minus#(false(),s(x),y) -> minus#(x,y)
p2: minus#(s(x),y) -> if_minus#(le(s(x),y),s(x),y)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_minus#_A(x1,x2,x3) = max{x1 + 2, x2 + 1, x3 + 4}
          false_A = 4
          s_A(x1) = x1 + 5
          minus#_A(x1,x2) = max{x1 + 6, x2 + 4}
          le_A(x1,x2) = max{x1 + 3, x2 - 6}
          |0|_A = 11
          true_A = 12
    
      precedence:
      
        if_minus# = s = minus# = le = |0| > false = true
      
      partial status:
    
        pi(if_minus#) = [2]
        pi(false) = []
        pi(s) = [1]
        pi(minus#) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_minus#_A(x1,x2,x3) = max{1, x2}
          false_A = 7
          s_A(x1) = x1 + 8
          minus#_A(x1,x2) = x1
          le_A(x1,x2) = 1
          |0|_A = 0
          true_A = 2
    
      precedence:
      
        true > if_minus# = false = s = minus# = le = |0|
      
      partial status:
    
        pi(if_minus#) = []
        pi(false) = []
        pi(s) = [1]
        pi(minus#) = [1]
        pi(le) = []
        pi(|0|) = []
        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: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)

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.