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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          quot#_A(x1,x2) = x1 + 1
          s_A(x1) = x1 + 4
          minus_A(x1,x2) = x1 + 2
          le_A(x1,x2) = x1 + x2 + 1
          |0|_A() = 6
          true_A() = 6
          false_A() = 3
          ifMinus_A(x1,x2,x3) = x2 + 2
  
    precedence:
    
      s = minus = |0| = true = ifMinus > false > quot# = le
    
    partial status:
  
      pi(quot#) = []
      pi(s) = []
      pi(minus) = []
      pi(le) = [1, 2]
      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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          ifMinus#_A(x1,x2,x3) = x2
          false_A() = 7
          s_A(x1) = x1 + 6
          minus#_A(x1,x2) = x1 + 5
          le_A(x1,x2) = x2 + 5
          |0|_A() = 3
          true_A() = 4
  
    precedence:
    
      ifMinus# = false = minus# = le = |0| = true > s
    
    partial status:
  
      pi(ifMinus#) = [2]
      pi(false) = []
      pi(s) = []
      pi(minus#) = [1]
      pi(le) = [2]
      pi(|0|) = []
      pi(true) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          le#_A(x1,x2) = x2
          s_A(x1) = x1 + 1
  
    precedence:
    
      le# = s
    
    partial status:
  
      pi(le#) = [2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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