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)
  pred(s(x)) -> x
  minus(x,|0|()) -> x
  minus(x,s(y)) -> pred(minus(x,y))
  mod(|0|(),y) -> |0|()
  mod(s(x),|0|()) -> |0|()
  mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
  if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
  if_mod(false(),s(x),s(y)) -> s(x)

-- SCC decomposition.

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

p1: le#(s(x),s(y)) -> le#(x,y)
p2: minus#(x,s(y)) -> pred#(minus(x,y))
p3: minus#(x,s(y)) -> minus#(x,y)
p4: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))
p5: mod#(s(x),s(y)) -> le#(y,x)
p6: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p7: if_mod#(true(),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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: mod(|0|(),y) -> |0|()
r8: mod(s(x),|0|()) -> |0|()
r9: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r10: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r11: if_mod(false(),s(x),s(y)) -> s(x)

The estimated dependency graph contains the following SCCs:

  {p4, p6}
  {p1}
  {p3}


-- Reduction pair.

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

p1: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p2: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: mod(|0|(),y) -> |0|()
r8: mod(s(x),|0|()) -> |0|()
r9: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r10: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r11: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      if_mod#_A(x1,x2,x3) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,1,0,0)) x2 + ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x3 + (0,0,1,0)
      true_A() = (1,1,1,1)
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,0)) x1 + (4,1,1,1)
      mod#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(1,0,1,0)) x1 + (0,3,0,7)
      minus_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,1,0,0),(0,0,0,0)) x1 + (1,1,1,1)
      le_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x2 + (2,2,2,2)
      pred_A(x1) = x1 + (0,0,2,2)
      |0|_A() = (1,1,1,1)
      false_A() = (0,0,0,0)

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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: mod(|0|(),y) -> |0|()
r8: mod(s(x),|0|()) -> |0|()
r9: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r10: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r11: if_mod(false(),s(x),s(y)) -> s(x)

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, r10, r11

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#(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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: mod(|0|(),y) -> |0|()
r8: mod(s(x),|0|()) -> |0|()
r9: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r10: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r11: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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