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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_mod#_A(x1,x2,x3) = max{8, x2 - 3, x3 - 7}
        true_A = 5
        s_A(x1) = x1 + 12
        mod#_A(x1,x2) = max{2, x1 - 3, x2 - 7}
        minus_A(x1,x2) = max{9, x1 + 2, x2 + 4}
        le_A(x1,x2) = max{8, x1 + 6, x2 + 4}
        |0|_A = 6
        false_A = 0
        pred_A(x1) = max{15, x1 - 10}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_mod#_A(x1,x2,x3) = max{0, x1 - 20, x2 - 38, x3 - 32}
          true_A = 0
          s_A(x1) = max{30, x1 + 23}
          mod#_A(x1,x2) = max{0, x1 - 33, x2 - 32}
          minus_A(x1,x2) = max{19, x1 + 1}
          le_A(x1,x2) = max{19, x1 - 31, x2 - 51}
          |0|_A = 9
          false_A = 0
          pred_A(x1) = max{2, x1 - 20}
          if_mod#!_A(x1,x2,x3) = max{11, x2 - 10}
          true!_A = 2
          s!_A(x1) = 12
          mod#!_A(x1,x2) = max{12, x1 - 2}
          minus!_A(x1,x2) = 12
          le!_A(x1,x2) = max{10, x1 + 1, x2 + 1}
          |0|!_A = 12
          false!_A = 11
          pred!_A(x1) = max{11, x1 - 21}
  
    partial status:
  
      pi(if_mod#) = []
      pi(true) = []
      pi(s) = [1]
      pi(mod#) = []
      pi(minus) = []
      pi(le) = [1]
      pi(|0|) = []
      pi(false) = []
      pi(pred) = []

The following rules are removed:

  p1, p2

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

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = 0
        s_A(x1) = max{3, x1 - 10}
        le_A(x1,x2) = max{4, x1 - 20, x2 - 19}
        |0|_A = 2
        true_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          le#_A(x1,x2) = max{x1, x2}
          s_A(x1) = max{12, x1 + 1}
          le_A(x1,x2) = max{8, x1 - 13, x2 - 3}
          |0|_A = 2
          true_A = 0
          le#!_A(x1,x2) = max{4, x1 - 2, x2 - 13}
          s!_A(x1) = max{3, x1 - 46}
          le!_A(x1,x2) = max{2, x1 - 25, x2 - 43}
          |0|!_A = 2
          true!_A = 1
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []

The following rules are removed:

  p1

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

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = x1 + 5
        s_A(x1) = max{3, x1 - 16}
        le_A(x1,x2) = max{x1 - 17, x2 + 15}
        |0|_A = 2
        true_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus#_A(x1,x2) = x2
          s_A(x1) = x1 + 12
          le_A(x1,x2) = max{13, x1 - 16, x2 - 16}
          |0|_A = 15
          true_A = 0
          minus#!_A(x1,x2) = max{9, x2 - 2}
          s!_A(x1) = max{8, x1 - 78}
          le!_A(x1,x2) = max{9, x1 - 10, x2 - 13}
          |0|!_A = 7
          true!_A = 6
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = []
      pi(le) = [2]
      pi(|0|) = []
      pi(true) = []

The following rules are removed:

  p1

Then no dependency pair remains.