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) -> ifMinus(le(s(X),Y),s(X),Y)
  ifMinus(true(),s(X),Y) -> |0|()
  ifMinus(false(),s(X),Y) -> s(minus(X,Y))
  quot(|0|(),s(Y)) -> |0|()
  quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(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#(s(X),Y) -> ifMinus#(le(s(X),Y),s(X),Y)
p3: minus#(s(X),Y) -> le#(s(X),Y)
p4: ifMinus#(false(),s(X),Y) -> minus#(X,Y)
p5: quot#(s(X),s(Y)) -> quot#(minus(X,Y),s(Y))
p6: quot#(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) -> ifMinus(le(s(X),Y),s(X),Y)
r6: ifMinus(true(),s(X),Y) -> |0|()
r7: ifMinus(false(),s(X),Y) -> s(minus(X,Y))
r8: quot(|0|(),s(Y)) -> |0|()
r9: quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))

The estimated dependency graph contains the following SCCs:

  {p5}
  {p2, p4}
  {p1}


-- Reduction pair.

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

p1: quot#(s(X),s(Y)) -> quot#(minus(X,Y),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) -> ifMinus(le(s(X),Y),s(X),Y)
r6: ifMinus(true(),s(X),Y) -> |0|()
r7: ifMinus(false(),s(X),Y) -> s(minus(X,Y))
r8: quot(|0|(),s(Y)) -> |0|()
r9: quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))

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:
        
          quot#_A(x1,x2) = x1 + 2
          s_A(x1) = x1 + 9
          minus_A(x1,x2) = x1 + 4
          le_A(x1,x2) = max{14, x1 + 6, x2 + 10}
          |0|_A = 8
          true_A = 7
          false_A = 0
          ifMinus_A(x1,x2,x3) = max{6, x2 + 4}
    
      precedence:
      
        quot# = minus = |0| > le = true = false = ifMinus > s
      
      partial status:
    
        pi(quot#) = [1]
        pi(s) = [1]
        pi(minus) = [1]
        pi(le) = [1, 2]
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(ifMinus) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          quot#_A(x1,x2) = max{0, x1 - 9}
          s_A(x1) = max{9, x1 + 4}
          minus_A(x1,x2) = x1 + 4
          le_A(x1,x2) = max{x1 + 14, x2 - 5}
          |0|_A = 2
          true_A = 15
          false_A = 3
          ifMinus_A(x1,x2,x3) = 0
    
      precedence:
      
        quot# = s = minus = le = |0| = true = false = ifMinus
      
      partial status:
    
        pi(quot#) = []
        pi(s) = [1]
        pi(minus) = [1]
        pi(le) = [1]
        pi(|0|) = []
        pi(true) = []
        pi(false) = []
        pi(ifMinus) = []
    

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: ifMinus#(false(),s(X),Y) -> minus#(X,Y)
p2: minus#(s(X),Y) -> ifMinus#(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) -> ifMinus(le(s(X),Y),s(X),Y)
r6: ifMinus(true(),s(X),Y) -> |0|()
r7: ifMinus(false(),s(X),Y) -> s(minus(X,Y))
r8: quot(|0|(),s(Y)) -> |0|()
r9: quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))

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:
        
          ifMinus#_A(x1,x2,x3) = max{x2 + 1, x3 + 8}
          false_A = 5
          s_A(x1) = x1 + 6
          minus#_A(x1,x2) = max{x1 + 1, x2 + 8}
          le_A(x1,x2) = max{x1 + 5, x2 - 7}
          |0|_A = 13
          true_A = 14
    
      precedence:
      
        ifMinus# = s = minus# = le = |0| > false = true
      
      partial status:
    
        pi(ifMinus#) = [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:
        
          ifMinus#_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 > ifMinus# = false = s = minus# = le = |0|
      
      partial status:
    
        pi(ifMinus#) = []
        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) -> ifMinus(le(s(X),Y),s(X),Y)
r6: ifMinus(true(),s(X),Y) -> |0|()
r7: ifMinus(false(),s(X),Y) -> s(minus(X,Y))
r8: quot(|0|(),s(Y)) -> |0|()
r9: quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(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.