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

  r4, r5, r6, r7

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_mod#_A(x1,x2,x3) = x3
              true_A() = 5
              s_A(x1) = 1
              mod#_A(x1,x2) = x2
              minus_A(x1,x2) = 3
              le_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 3
              if_minus_A(x1,x2,x3) = x2 + 2
              false_A() = 6
              if_mod#!_A(x1,x2,x3) = 0
              true!_A() = 2
              s!_A(x1) = 4
              mod#!_A(x1,x2) = 1
              minus!_A(x1,x2) = 4
              le!_A(x1,x2) = x1 + x2
              |0|!_A() = 3
              if_minus!_A(x1,x2,x3) = 4
              false!_A() = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_mod#_A(x1,x2,x3) = max{30, x3 + 2}
            true_A = 1
            s_A(x1) = x1 + 27
            mod#_A(x1,x2) = max{30, x2 + 2}
            minus_A(x1,x2) = x1 + 24
            le_A(x1,x2) = x2 + 24
            |0|_A = 9
            if_minus_A(x1,x2,x3) = x2 + 24
            false_A = 3
            if_mod#!_A(x1,x2,x3) = max{26, x2 + 6, x3 - 28}
            true!_A = 1
            s!_A(x1) = max{24, x1}
            mod#!_A(x1,x2) = 25
            minus!_A(x1,x2) = max{0, x1 - 1}
            le!_A(x1,x2) = max{2, x1 - 9}
            |0|!_A = 9
            if_minus!_A(x1,x2,x3) = max{22, x2 - 2}
            false!_A = 10
        
  
    partial status:
  
      pi(if_mod#) = [2]
      pi(true) = []
      pi(s) = [1]
      pi(mod#) = [1]
      pi(minus) = []
      pi(le) = [1]
      pi(|0|) = []
      pi(if_minus) = []
      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: 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:
  
      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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_minus#_A(x1,x2,x3) = x2
              false_A() = 1
              s_A(x1) = x1 + 2
              minus#_A(x1,x2) = x1
              le_A(x1,x2) = 1
              |0|_A() = 3
              true_A() = 1
              minus_A(x1,x2) = x1 + x2 + 1
              if_minus#!_A(x1,x2,x3) = x2 + 2
              false!_A() = 2
              s!_A(x1) = 2
              minus#!_A(x1,x2) = x1 + 3
              le!_A(x1,x2) = 2
              |0|!_A() = 0
              true!_A() = 1
              minus!_A(x1,x2) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_minus#_A(x1,x2,x3) = max{0, x1 - 14, x2 - 15, x3 - 4}
            false_A = 1
            s_A(x1) = max{12, x1 + 2}
            minus#_A(x1,x2) = max{0, x1 - 13, x2 - 4}
            le_A(x1,x2) = 12
            |0|_A = 0
            true_A = 10
            minus_A(x1,x2) = max{4, x1 - 4}
            if_minus#!_A(x1,x2,x3) = max{34, x2 - 10}
            false!_A = 23
            s!_A(x1) = max{33, x1 + 24}
            minus#!_A(x1,x2) = max{x1 + 33, x2 + 35}
            le!_A(x1,x2) = max{33, x2 + 22}
            |0|!_A = 20
            true!_A = 21
            minus!_A(x1,x2) = 18
        
  
    partial status:
  
      pi(if_minus#) = []
      pi(false) = []
      pi(s) = [1]
      pi(minus#) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = []

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

  (no rules)

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              le#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 7
              le#!_A(x1,x2) = x1
              s!_A(x1) = x1 + 7
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{14, x1 - 30}
            le#!_A(x1,x2) = max{0, x1 - 15}
            s!_A(x1) = max{18, x1 + 4}
        
  
    partial status:
  
      pi(le#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.