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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      quot#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2
      s_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (5,2,1,1)
      minus_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + (2,1,0,0)
      le_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x2 + (2,2,2,2)
      |0|_A() = (6,6,5,7)
      true_A() = (1,1,1,3)
      false_A() = (1,15,1,1)
      ifMinus_A(x1,x2,x3) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,0)) x2 + (2,0,1,2)

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      ifMinus#_A(x1,x2,x3) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x3 + (0,0,1,0)
      false_A() = (1,1,2,2)
      s_A(x1) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(1,1,1,0)) x1 + (2,0,1,1)
      minus#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (0,1,0,1)
      le_A(x1,x2) = (2,2,1,1)
      |0|_A() = (1,1,1,1)
      true_A() = (0,3,0,2)

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 monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      le#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,0,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,0,1,1)) x2
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,1,1,1)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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