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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      div#_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(0,1,0),(0,1,1)) x2
      s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (2,1,1)
      minus_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((0,0,0),(1,0,0),(1,1,0)) x2 + (1,3,1)
      |0|_A() = (1,1,1)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      minus#_A(x1,x2) = x1
      s_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (1,1,1)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      geq#_A(x1,x2) = x1
      s_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1

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