YES

We show the termination of the TRS R:

  minus(|0|(),Y) -> |0|()
  minus(s(X),s(Y)) -> minus(X,Y)
  geq(X,|0|()) -> true()
  geq(|0|(),s(Y)) -> false()
  geq(s(X),s(Y)) -> geq(X,Y)
  div(|0|(),s(Y)) -> |0|()
  div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|())
  if(true(),X,Y) -> X
  if(false(),X,Y) -> Y

-- SCC decomposition.

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

p1: minus#(s(X),s(Y)) -> minus#(X,Y)
p2: geq#(s(X),s(Y)) -> geq#(X,Y)
p3: div#(s(X),s(Y)) -> if#(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|())
p4: div#(s(X),s(Y)) -> geq#(X,Y)
p5: div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))
p6: div#(s(X),s(Y)) -> minus#(X,Y)

and R consists of:

r1: minus(|0|(),Y) -> |0|()
r2: minus(s(X),s(Y)) -> minus(X,Y)
r3: geq(X,|0|()) -> true()
r4: geq(|0|(),s(Y)) -> false()
r5: geq(s(X),s(Y)) -> geq(X,Y)
r6: div(|0|(),s(Y)) -> |0|()
r7: div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|())
r8: if(true(),X,Y) -> X
r9: if(false(),X,Y) -> Y

The estimated dependency graph contains the following SCCs:

  {p5}
  {p1}
  {p2}


-- Reduction pair.

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

p1: div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))

and R consists of:

r1: minus(|0|(),Y) -> |0|()
r2: minus(s(X),s(Y)) -> minus(X,Y)
r3: geq(X,|0|()) -> true()
r4: geq(|0|(),s(Y)) -> false()
r5: geq(s(X),s(Y)) -> geq(X,Y)
r6: div(|0|(),s(Y)) -> |0|()
r7: div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|())
r8: if(true(),X,Y) -> X
r9: if(false(),X,Y) -> Y

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > minus > s > div#
      
      argument filter:
    
        pi(div#) = [1]
        pi(s) = [1]
        pi(minus) = 1
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        div#_A(x1,x2) = x1
        s_A(x1) = x1 + 1
        minus_A(x1,x2) = 2
        |0|_A() = 1
    

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: minus#(s(X),s(Y)) -> minus#(X,Y)

and R consists of:

r1: minus(|0|(),Y) -> |0|()
r2: minus(s(X),s(Y)) -> minus(X,Y)
r3: geq(X,|0|()) -> true()
r4: geq(|0|(),s(Y)) -> false()
r5: geq(s(X),s(Y)) -> geq(X,Y)
r6: div(|0|(),s(Y)) -> |0|()
r7: div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|())
r8: if(true(),X,Y) -> X
r9: if(false(),X,Y) -> Y

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > minus#
      
      argument filter:
    
        pi(minus#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        minus#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

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: geq#(s(X),s(Y)) -> geq#(X,Y)

and R consists of:

r1: minus(|0|(),Y) -> |0|()
r2: minus(s(X),s(Y)) -> minus(X,Y)
r3: geq(X,|0|()) -> true()
r4: geq(|0|(),s(Y)) -> false()
r5: geq(s(X),s(Y)) -> geq(X,Y)
r6: div(|0|(),s(Y)) -> |0|()
r7: div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|())
r8: if(true(),X,Y) -> X
r9: if(false(),X,Y) -> Y

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > geq#
      
      argument filter:
    
        pi(geq#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        geq#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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