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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if_mod#_A(x1,x2,x3) = x2 + 1
          true_A() = 1
          s_A(x1) = x1 + 5
          mod#_A(x1,x2) = x1 + 2
          minus_A(x1,x2) = x1 + 2
          le_A(x1,x2) = 4
          |0|_A() = 2
          false_A() = 1
          pred_A(x1) = x1
  
    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 - 19, x2 - 15, x3 - 11}
          true_A = 4
          s_A(x1) = max{9, x1 + 4}
          mod#_A(x1,x2) = max{0, x1 - 12, x2 - 11}
          minus_A(x1,x2) = max{2, x1 + 1, x2 - 11}
          le_A(x1,x2) = max{18, x1 - 11, x2 - 3}
          |0|_A = 2
          false_A = 7
          pred_A(x1) = max{2, x1 - 3}
          if_mod#!_A(x1,x2,x3) = max{x1 + 2, x2 - 3, x3 - 3}
          true!_A = 1
          s!_A(x1) = 9
          mod#!_A(x1,x2) = 7
          minus!_A(x1,x2) = max{11, x1 - 1, x2 - 10}
          le!_A(x1,x2) = max{8, x1 - 19, x2 - 10}
          |0|!_A = 2
          false!_A = 3
          pred!_A(x1) = max{11, x1 - 9}
  
    partial status:
  
      pi(if_mod#) = []
      pi(true) = []
      pi(s) = []
      pi(mod#) = []
      pi(minus) = [1]
      pi(le) = []
      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, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          le#_A(x1,x2) = x1 + x2 + 4
          s_A(x1) = x1 + 3
          le_A(x1,x2) = x1 + x2 + 4
          |0|_A() = 2
          true_A() = 1
          false_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{11, x1 + 1}
          le_A(x1,x2) = max{8, x1 - 12, x2 - 3}
          |0|_A = 2
          true_A = 0
          false_A = 0
          le#!_A(x1,x2) = max{6, x1 - 12, x2 - 12}
          s!_A(x1) = max{5, x1 + 2}
          le!_A(x1,x2) = max{4, x1 - 15, x2 - 12}
          |0|!_A = 4
          true!_A = 1
          false!_A = 3
  
    partial status:
  
      pi(le#) = [1]
      pi(s) = [1]
      pi(le) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []

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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          minus#_A(x1,x2) = x2
          s_A(x1) = x1 + 3
          le_A(x1,x2) = x1 + x2 + 4
          |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) = max{9, x1}
          le_A(x1,x2) = max{6, x1 - 11, x2 - 3}
          |0|_A = 2
          true_A = 0
          minus#!_A(x1,x2) = max{5, x1 - 1, x2 - 10}
          s!_A(x1) = max{4, x1 - 36}
          le!_A(x1,x2) = max{5, x1 - 28, x2 - 11}
          |0|!_A = 2
          true!_A = 1
  
    partial status:
  
      pi(minus#) = [2]
      pi(s) = []
      pi(le) = [2]
      pi(|0|) = []
      pi(true) = []

The following rules are removed:

  p1

Then no dependency pair remains.