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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_mod#_A(x1,x2,x3) = max{1, x3 - 28}
            true_A = 59
            s_A(x1) = max{27, x1 + 12}
            mod#_A(x1,x2) = max{1, x2 - 28}
            minus_A(x1,x2) = x1 + 8
            le_A(x1,x2) = x2 + 59
            |0|_A = 6
            false_A = 1
            if_mod#!_A(x1,x2,x3) = max{32, x1 - 25, x2 + 22}
            true!_A = 1
            s!_A(x1) = max{33, x1 + 31}
            mod#!_A(x1,x2) = x1 + 22
            minus!_A(x1,x2) = max{30, x1 - 13}
            le!_A(x1,x2) = 2
            |0|!_A = 30
            false!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_mod#_A(x1,x2,x3) = max{1, x3 - 49}
            true_A = 0
            s_A(x1) = max{48, x1 + 13}
            mod#_A(x1,x2) = max{1, x2 - 49}
            minus_A(x1,x2) = max{47, x1 + 13, x2 + 13}
            le_A(x1,x2) = max{33, x1 - 29, x2 - 31}
            |0|_A = 28
            false_A = 5
            if_mod#!_A(x1,x2,x3) = max{63, x1 + 30, x2 + 15, x3 + 15}
            true!_A = 58
            s!_A(x1) = 61
            mod#!_A(x1,x2) = max{29, x1 + 15, x2 + 15}
            minus!_A(x1,x2) = max{64, x2 - 49}
            le!_A(x1,x2) = 59
            |0|!_A = 2
            false!_A = 3
        
  
    partial status:
  
      pi(if_mod#) = [1, 2]
      pi(true) = []
      pi(s) = [1]
      pi(mod#) = [1]
      pi(minus) = [1]
      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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = max{x1, x2}
            s_A(x1) = max{8, x1 + 6}
            le_A(x1,x2) = max{10, x1 - 9, x2 - 5}
            |0|_A = 2
            true_A = 0
            le#!_A(x1,x2) = max{4, x1 - 24, x2 - 9}
            s!_A(x1) = max{3, x1 - 23}
            le!_A(x1,x2) = max{4, x1 - 9, x2 - 9}
            |0|!_A = 2
            true!_A = 1
        
        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{3, x1}
            le_A(x1,x2) = max{2, x1 - 4, x2 - 3}
            |0|_A = 2
            true_A = 0
            le#!_A(x1,x2) = max{1, x1 - 10, x2 - 4}
            s!_A(x1) = max{3, x1 - 9}
            le!_A(x1,x2) = max{7, x1 - 4, x2 - 4}
            |0|!_A = 2
            true!_A = 6
        
  
    partial status:
  
      pi(le#) = [1]
      pi(s) = [1]
      pi(le) = [1]
      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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{x1, x2}
            s_A(x1) = max{7, x1 + 2}
            le_A(x1,x2) = max{9, x1 - 8, x2 - 3}
            |0|_A = 2
            true_A = 0
            minus#!_A(x1,x2) = max{4, x1 - 8, x2 - 8}
            s!_A(x1) = max{3, x1 - 26}
            le!_A(x1,x2) = max{4, x1 - 8, x2 - 8}
            |0|!_A = 2
            true!_A = 1
        
        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{2, x1}
            le_A(x1,x2) = max{2, x1 - 3, x2 - 3}
            |0|_A = 2
            true_A = 0
            minus#!_A(x1,x2) = max{4, x1 - 9, x2 - 10}
            s!_A(x1) = max{3, x1 - 9}
            le!_A(x1,x2) = max{5, x1 - 3, x2 - 3}
            |0|!_A = 2
            true!_A = 4
        
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]
      pi(le) = [1]
      pi(|0|) = []
      pi(true) = []

The following rules are removed:

  p1

Then no dependency pair remains.