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

The estimated dependency graph contains the following SCCs:

  {p4, p6}
  {p1}
  {p3}


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

The set of usable rules consists of

  r1, r2, r3, r4, r5, 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_mod#_A(x1,x2,x3) = x3
              true_A() = 1
              s_A(x1) = x1 + 1
              mod#_A(x1,x2) = x2
              minus_A(x1,x2) = x1 + 1
              le_A(x1,x2) = x1 + 1
              |0|_A() = 2
              false_A() = 1
              pred_A(x1) = x1
              if_mod#!_A(x1,x2,x3) = 3
              true!_A() = 4
              s!_A(x1) = 2
              mod#!_A(x1,x2) = 3
              minus!_A(x1,x2) = 1
              le!_A(x1,x2) = 1
              |0|!_A() = 5
              false!_A() = 0
              pred!_A(x1) = 0
        
        2. 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() = 1
              s_A(x1) = x1 + 3
              mod#_A(x1,x2) = x2
              minus_A(x1,x2) = x1 + 1
              le_A(x1,x2) = x1 + 1
              |0|_A() = 2
              false_A() = 1
              pred_A(x1) = x1
              if_mod#!_A(x1,x2,x3) = 4
              true!_A() = 5
              s!_A(x1) = 3
              mod#!_A(x1,x2) = 4
              minus!_A(x1,x2) = x2 + 2
              le!_A(x1,x2) = 2
              |0|!_A() = 6
              false!_A() = 7
              pred!_A(x1) = 4
        
  
    partial status:
  
      pi(if_mod#) = [2]
      pi(true) = []
      pi(s) = [1]
      pi(mod#) = [1]
      pi(minus) = [1]
      pi(le) = [1]
      pi(|0|) = []
      pi(false) = []
      pi(pred) = [1]

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

The set of usable rules consists of

  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:
              le#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              mod_A(x1,x2) = x1
              |0|_A() = 7
              le#!_A(x1,x2) = x1 + x2 + 3
              s!_A(x1) = 2
              mod!_A(x1,x2) = x1 + x2 + 2
              |0|!_A() = 8
        
        2. 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 + 6
              mod_A(x1,x2) = x1
              |0|_A() = 10
              le#!_A(x1,x2) = x1 + x2 + 11
              s!_A(x1) = x1 + 10
              mod!_A(x1,x2) = x1 + x2
              |0|!_A() = 9
        
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(mod) = []
      pi(|0|) = []

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

The set of usable rules consists of

  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:
              minus#_A(x1,x2) = x2
              s_A(x1) = x1 + 1
              mod_A(x1,x2) = x1
              |0|_A() = 2
              minus#!_A(x1,x2) = x2
              s!_A(x1) = 2
              mod!_A(x1,x2) = 3
              |0|!_A() = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              minus#_A(x1,x2) = x2
              s_A(x1) = x1 + 1
              mod_A(x1,x2) = x1
              |0|_A() = 5
              minus#!_A(x1,x2) = x2 + 5
              s!_A(x1) = x1 + 4
              mod!_A(x1,x2) = 2
              |0|!_A() = 1
        
  
    partial status:
  
      pi(minus#) = [2]
      pi(s) = [1]
      pi(mod) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.