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:
  
      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, x1 - 23, x3 - 32}
            true_A = 6
            s_A(x1) = x1 + 31
            mod#_A(x1,x2) = max{1, x2 - 32}
            minus_A(x1,x2) = x1
            le_A(x1,x2) = max{23, x1 + 1}
            |0|_A = 13
            false_A = 32
            if_minus_A(x1,x2,x3) = max{13, x1 - 1, x2}
            if_mod#!_A(x1,x2,x3) = 8
            true!_A = 5
            s!_A(x1) = x1 + 19
            mod#!_A(x1,x2) = 8
            minus!_A(x1,x2) = x1 + 17
            le!_A(x1,x2) = max{6, x1 - 29}
            |0|!_A = 4
            false!_A = 3
            if_minus!_A(x1,x2,x3) = max{5, x1 - 24, x2 - 15}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_mod#_A(x1,x2,x3) = max{24, x1 + 16}
            true_A = 2
            s_A(x1) = 3
            mod#_A(x1,x2) = 24
            minus_A(x1,x2) = x1
            le_A(x1,x2) = 8
            |0|_A = 0
            false_A = 2
            if_minus_A(x1,x2,x3) = max{1, x1 - 9, x2}
            if_mod#!_A(x1,x2,x3) = x2 + 1
            true!_A = 8
            s!_A(x1) = x1 + 6
            mod#!_A(x1,x2) = max{4, x2 + 1}
            minus!_A(x1,x2) = 0
            le!_A(x1,x2) = 7
            |0|!_A = 6
            false!_A = 7
            if_minus!_A(x1,x2,x3) = max{4, x1 - 1}
        
  
    partial status:
  
      pi(if_mod#) = [2]
      pi(true) = []
      pi(s) = []
      pi(mod#) = [1]
      pi(minus) = []
      pi(le) = []
      pi(|0|) = []
      pi(false) = []
      pi(if_minus) = [2]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 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

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_minus#_A(x1,x2,x3) = max{0, x1 - 3, x2 - 30, x3 - 11}
            false_A = 2
            s_A(x1) = max{20, x1 + 5}
            minus#_A(x1,x2) = max{0, x1 - 30, x2 - 11}
            le_A(x1,x2) = max{2, x1 - 27, x2 - 9}
            |0|_A = 8
            true_A = 0
            if_minus#!_A(x1,x2,x3) = max{6, x1 - 18, x2 - 42, x3 - 20}
            false!_A = 3
            s!_A(x1) = max{5, x1 - 40}
            minus#!_A(x1,x2) = max{6, x1 - 41, x2 - 20}
            le!_A(x1,x2) = max{5, x1 - 45, x2 - 21}
            |0|!_A = 2
            true!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_minus#_A(x1,x2,x3) = max{0, x1 - 15, x2 - 43, x3 - 17}
            false_A = 14
            s_A(x1) = max{27, x1 + 4}
            minus#_A(x1,x2) = max{0, x1 - 43, x2 - 17}
            le_A(x1,x2) = max{14, x1 - 28, x2 - 20}
            |0|_A = 19
            true_A = 0
            if_minus#!_A(x1,x2,x3) = max{4, x1 - 67, x2 - 97, x3 - 69}
            false!_A = 4
            s!_A(x1) = max{4, x1 - 90}
            minus#!_A(x1,x2) = max{4, x1 - 94, x2 - 69}
            le!_A(x1,x2) = max{3, x1 - 118, x2 - 91}
            |0|!_A = 3
            true!_A = 2
        
  
    partial status:
  
      pi(if_minus#) = [2]
      pi(false) = []
      pi(s) = [1]
      pi(minus#) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []

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

  r1, r2, r4, r6

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) = 0
            s_A(x1) = max{1, x1 - 9}
            le_A(x1,x2) = max{13, x1 - 4}
            |0|_A = 3
            true_A = 1
            false_A = 10
            minus_A(x1,x2) = max{8, x1 - 7, x2 - 7}
            if_minus_A(x1,x2,x3) = max{3, x1 - 12, x2 - 7, x3 - 7}
            le#!_A(x1,x2) = 0
            s!_A(x1) = 13
            le!_A(x1,x2) = 2
            |0|!_A = 6
            true!_A = 0
            false!_A = 1
            minus!_A(x1,x2) = x2 + 18
            if_minus!_A(x1,x2,x3) = max{x1 + 4, x3 + 18}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = max{x1, x2}
            s_A(x1) = x1 + 31
            le_A(x1,x2) = max{14, x1 - 16}
            |0|_A = 52
            true_A = 1
            false_A = 15
            minus_A(x1,x2) = max{64, x2 + 48}
            if_minus_A(x1,x2,x3) = max{x1 + 50, x2 + 33, x3 - 1}
            le#!_A(x1,x2) = 0
            s!_A(x1) = max{67, x1 + 5}
            le!_A(x1,x2) = 3
            |0|!_A = 69
            true!_A = 70
            false!_A = 1
            minus!_A(x1,x2) = 4
            if_minus!_A(x1,x2,x3) = max{68, x1 - 13}
        
  
    partial status:
  
      pi(le#) = [1]
      pi(s) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(minus) = []
      pi(if_minus) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.