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

The estimated dependency graph contains the following SCCs:

  {p3, p5}
  {p1}
  {p2}


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

The set of usable rules consists of

  r1, r2, r3, r4, r5

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 + 2
          true_A() = 4
          s_A(x1) = x1 + 2
          mod#_A(x1,x2) = x1 + 2
          minus_A(x1,x2) = x1 + 1
          le_A(x1,x2) = 7
          |0|_A() = 5
          false_A() = 1
  
    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 - 5, x2 - 34, x3 - 18}
          true_A = 3
          s_A(x1) = x1 + 13
          mod#_A(x1,x2) = max{0, x1 - 27, x2 - 18}
          minus_A(x1,x2) = max{26, x1 + 6, x2 + 11}
          le_A(x1,x2) = max{1, x1 - 14}
          |0|_A = 17
          false_A = 1
          if_mod#!_A(x1,x2,x3) = max{4, x1 + 1, x3 - 9}
          true!_A = 2
          s!_A(x1) = x1 + 11
          mod#!_A(x1,x2) = max{5, x2 - 3}
          minus!_A(x1,x2) = max{1, x1 - 25, x2 - 14}
          le!_A(x1,x2) = max{12, x2 - 18}
          |0|!_A = 4
          false!_A = 1
  
    partial status:
  
      pi(if_mod#) = []
      pi(true) = []
      pi(s) = []
      pi(mod#) = []
      pi(minus) = []
      pi(le) = []
      pi(|0|) = []
      pi(false) = []

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: minus(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: 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:
          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
  
    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{15, x1 + 9}
          le_A(x1,x2) = max{2, x1 - 16, x2 - 3}
          |0|_A = 2
          true_A = 0
          le#!_A(x1,x2) = max{1, x1 - 33, x2 - 16}
          s!_A(x1) = max{3, x1 - 32}
          le!_A(x1,x2) = max{4, x1 - 25, x2 - 10}
          |0|!_A = 2
          true!_A = 1
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = []
      pi(le) = [2]
      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#(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: minus(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: 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) = x1 + x2 + 4
          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) = max{x1, x2}
          s_A(x1) = max{8, x1 + 5}
          le_A(x1,x2) = max{2, x1 - 9, x2 - 3}
          |0|_A = 2
          true_A = 0
          minus#!_A(x1,x2) = max{1, x1 - 24, x2 - 23}
          s!_A(x1) = max{3, x1 - 23}
          le!_A(x1,x2) = max{4, x1 - 18, x2 - 7}
          |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.