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(|0|(),y) -> |0|()
  minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
  if_minus(true(),s(x),y) -> |0|()
  if_minus(false(),s(x),y) -> s(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),y) -> if_minus#(le(s(x),y),s(x),y)
p3: minus#(s(x),y) -> le#(s(x),y)
p4: if_minus#(false(),s(x),y) -> minus#(x,y)
p5: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))
p6: mod#(s(x),s(y)) -> le#(y,x)
p7: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p8: 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(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)

The estimated dependency graph contains the following SCCs:

  {p5, p7}
  {p2, p4}
  {p1}


-- 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(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

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 + x3
          true_A() = 1
          s_A(x1) = x1 + 3
          mod#_A(x1,x2) = x1 + x2 + 1
          minus_A(x1,x2) = x1
          le_A(x1,x2) = x2 + 2
          |0|_A() = 0
          false_A() = 1
          if_minus_A(x1,x2,x3) = x2
  
    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 - 12, x2 - 29, x3 - 28}
          true_A = 0
          s_A(x1) = max{22, x1}
          mod#_A(x1,x2) = max{0, x1 - 29, x2 - 28}
          minus_A(x1,x2) = max{28, x1 - 29, x2 - 20}
          le_A(x1,x2) = max{11, x1 - 29, x2 - 29}
          |0|_A = 28
          false_A = 11
          if_minus_A(x1,x2,x3) = max{28, x1 - 12, x2 - 29, x3 - 20}
          if_mod#!_A(x1,x2,x3) = max{5, x1 - 12, x2 - 23, x3 - 23}
          true!_A = 1
          s!_A(x1) = max{0, x1 - 26}
          mod#!_A(x1,x2) = 0
          minus!_A(x1,x2) = max{4, x1 - 24, x2 - 30}
          le!_A(x1,x2) = max{5, x1 - 47, x2 - 29}
          |0|!_A = 0
          false!_A = 0
          if_minus!_A(x1,x2,x3) = max{3, x1 - 42, x2 - 53, x3 - 31}
  
    partial status:
  
      pi(if_mod#) = []
      pi(true) = []
      pi(s) = [1]
      pi(mod#) = []
      pi(minus) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(false) = []
      pi(if_minus) = [2]

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: if_minus#(false(),s(x),y) -> minus#(x,y)
p2: minus#(s(x),y) -> if_minus#(le(s(x),y),s(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(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  r1, r4

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_minus#_A(x1,x2,x3) = 2
          false_A() = 5
          s_A(x1) = 4
          minus#_A(x1,x2) = 2
          le_A(x1,x2) = 1
          |0|_A() = 1
          true_A() = 0
          minus_A(x1,x2) = x1 + x2 + 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_minus#_A(x1,x2,x3) = max{1, x2 - 6, x3 - 7}
          false_A = 7
          s_A(x1) = x1 + 5
          minus#_A(x1,x2) = max{1, x1 - 6, x2 - 7}
          le_A(x1,x2) = max{7, x1 - 8, x2 - 3}
          |0|_A = 2
          true_A = 0
          minus_A(x1,x2) = max{10, x1 - 22, x2 - 24}
          if_minus#!_A(x1,x2,x3) = x2 + 1
          false!_A = 8
          s!_A(x1) = max{6, x1 + 4}
          minus#!_A(x1,x2) = x1 + 2
          le!_A(x1,x2) = max{3, x1 - 6, x2 - 15}
          |0|!_A = 3
          true!_A = 8
          minus!_A(x1,x2) = 7
  
    partial status:
  
      pi(if_minus#) = []
      pi(false) = []
      pi(s) = []
      pi(minus#) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = [1, 2]

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(|0|(),y) -> |0|()
r5: minus(s(x),y) -> if_minus(le(s(x),y),s(x),y)
r6: if_minus(true(),s(x),y) -> |0|()
r7: if_minus(false(),s(x),y) -> s(minus(x,y))
r8: mod(|0|(),y) -> |0|()
r9: mod(s(x),|0|()) -> |0|()
r10: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r11: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r12: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  r1, r4

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 + 5
          s_A(x1) = x1 + 4
          le_A(x1,x2) = x1 + x2 + 10
          |0|_A() = 2
          true_A() = 3
          minus_A(x1,x2) = x1 + x2 + 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{8, x1 - 17}
          le_A(x1,x2) = max{2, x1 - 9, x2 - 9}
          |0|_A = 2
          true_A = 0
          minus_A(x1,x2) = max{16, x1 - 15, x2 - 15}
          le#!_A(x1,x2) = max{5, x1 - 9, x2 - 9}
          s!_A(x1) = max{4, x1 - 114}
          le!_A(x1,x2) = max{6, x1 - 3, x2 - 9}
          |0|!_A = 2
          true!_A = 3
          minus!_A(x1,x2) = max{4, x1 - 60, x2 - 112}
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.