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:
  
      max/plus interpretation on natural numbers:
      
        if_mod#_A(x1,x2,x3) = max{17, x1 - 8, x2 + 13, x3 - 12}
        true_A = 25
        s_A(x1) = 11
        mod#_A(x1,x2) = max{x1 + 8, x2 + 13}
        minus_A(x1,x2) = 16
        le_A(x1,x2) = 26
        |0|_A = 13
        false_A = 12
        if_minus_A(x1,x2,x3) = max{12, x1 - 11}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_mod#_A(x1,x2,x3) = max{11, x3 - 19}
          true_A = 27
          s_A(x1) = x1 + 29
          mod#_A(x1,x2) = max{11, x2 - 19}
          minus_A(x1,x2) = x1 + 28
          le_A(x1,x2) = x2 + 27
          |0|_A = 49
          false_A = 1
          if_minus_A(x1,x2,x3) = max{49, x2 + 28}
          if_mod#!_A(x1,x2,x3) = max{24, x1 - 1, x2 - 3}
          true!_A = 0
          s!_A(x1) = 23
          mod#!_A(x1,x2) = max{25, x1 - 2}
          minus!_A(x1,x2) = 21
          le!_A(x1,x2) = 19
          |0|!_A = 31
          false!_A = 20
          if_minus!_A(x1,x2,x3) = x1 + 23
  
    partial status:
  
      pi(if_mod#) = []
      pi(true) = []
      pi(s) = []
      pi(mod#) = []
      pi(minus) = []
      pi(le) = []
      pi(|0|) = []
      pi(false) = []
      pi(if_minus) = []

The following rules are removed:

  p2



-- SCC decomposition.

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))

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:

  (no SCCs)

-- 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, r2, r3, r4

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_minus#_A(x1,x2,x3) = max{13, x1 - 8, x2 + 1, x3 - 22}
        false_A = 1
        s_A(x1) = max{11, x1 + 5}
        minus#_A(x1,x2) = max{13, x1 + 1, x2 - 22}
        le_A(x1,x2) = max{7, x1 - 24, x2 - 23}
        |0|_A = 2
        true_A = 3
        minus_A(x1,x2) = max{8, x1 - 17, x2 - 22}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_minus#_A(x1,x2,x3) = max{0, x1 - 21, x2 - 34, x3 - 24}
          false_A = 2
          s_A(x1) = max{19, x1 + 10}
          minus#_A(x1,x2) = max{0, x1 - 34, x2 - 24}
          le_A(x1,x2) = max{20, x1 - 21, x2 - 3}
          |0|_A = 2
          true_A = 0
          minus_A(x1,x2) = max{15, x1 - 34, x2 - 25}
          if_minus#!_A(x1,x2,x3) = max{x1 - 42, x2 - 52, x3 + 3}
          false!_A = 1
          s!_A(x1) = max{4, x1 - 43}
          minus#!_A(x1,x2) = max{x1 - 42, x2 + 3}
          le!_A(x1,x2) = max{2, x1 - 63, x2 - 71}
          |0|!_A = 6
          true!_A = 7
          minus!_A(x1,x2) = max{8, x1 - 55, x2 - 70}
  
    partial status:
  
      pi(if_minus#) = [2]
      pi(false) = []
      pi(s) = []
      pi(minus#) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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

  r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = max{8, x1}
        s_A(x1) = x1 + 8
        le_A(x1,x2) = max{14, x1 - 9, x2 - 9}
        |0|_A = 3
        false_A = 1
  
    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{2, x1 - 12}
          le_A(x1,x2) = max{0, x1 - 5, x2 - 1}
          |0|_A = 9
          false_A = 1
          le#!_A(x1,x2) = max{15, x1 + 13}
          s!_A(x1) = max{28, x1 + 14}
          le!_A(x1,x2) = max{29, x1 - 10, x2 - 10}
          |0|!_A = 25
          false!_A = 23
  
    partial status:
  
      pi(le#) = [1]
      pi(s) = []
      pi(le) = []
      pi(|0|) = []
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.