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)
  gcd(|0|(),y) -> y
  gcd(s(x),|0|()) -> s(x)
  gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
  if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
  if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),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: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),s(y))
p4: gcd#(s(x),s(y)) -> le#(y,x)
p5: if_gcd#(true(),s(x),s(y)) -> gcd#(minus(x,y),s(y))
p6: if_gcd#(true(),s(x),s(y)) -> minus#(x,y)
p7: if_gcd#(false(),s(x),s(y)) -> gcd#(minus(y,x),s(x))
p8: if_gcd#(false(),s(x),s(y)) -> minus#(y,x)

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: gcd(|0|(),y) -> y
r7: gcd(s(x),|0|()) -> s(x)
r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x))

The estimated dependency graph contains the following SCCs:

  {p3, p5, p7}
  {p1}
  {p2}


-- Reduction pair.

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

p1: if_gcd#(false(),s(x),s(y)) -> gcd#(minus(y,x),s(x))
p2: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),s(y))
p3: if_gcd#(true(),s(x),s(y)) -> gcd#(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(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: gcd(|0|(),y) -> y
r7: gcd(s(x),|0|()) -> s(x)
r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),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_gcd#_A(x1,x2,x3) = max{0, x1 - 7, x2 - 56, x3 - 57}
            false_A = 6
            s_A(x1) = max{45, x1 + 29}
            gcd#_A(x1,x2) = max{0, x1 - 53, x2 - 57}
            minus_A(x1,x2) = max{43, x1 + 7, x2 - 30}
            le_A(x1,x2) = max{0, x1 - 39, x2 - 39}
            true_A = 0
            |0|_A = 38
            if_gcd#!_A(x1,x2,x3) = max{x1 + 21, x2 - 26, x3 - 46}
            false!_A = 2
            s!_A(x1) = 29
            gcd#!_A(x1,x2) = max{21, x1 - 25, x2 - 46}
            minus!_A(x1,x2) = 28
            le!_A(x1,x2) = max{3, x1 - 46, x2 - 46}
            true!_A = 2
            |0|!_A = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_gcd#_A(x1,x2,x3) = max{0, x1 - 29, x2 - 49, x3 - 50}
            false_A = 20
            s_A(x1) = max{46, x1 + 28}
            gcd#_A(x1,x2) = max{0, x1 - 47, x2 - 50}
            minus_A(x1,x2) = max{45, x1 + 1}
            le_A(x1,x2) = max{19, x1 - 27, x2 - 29}
            true_A = 24
            |0|_A = 57
            if_gcd#!_A(x1,x2,x3) = max{9, x2 - 21}
            false!_A = 4
            s!_A(x1) = x1 + 6
            gcd#!_A(x1,x2) = max{25, x1 - 20}
            minus!_A(x1,x2) = max{6, x1 - 47}
            le!_A(x1,x2) = max{5, x1 - 68, x2 - 65}
            true!_A = 2
            |0|!_A = 3
        
  
    partial status:
  
      pi(if_gcd#) = [1]
      pi(false) = []
      pi(s) = [1]
      pi(gcd#) = []
      pi(minus) = [1]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []

The following rules are removed:

  p1, p2



-- SCC decomposition.

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

p1: if_gcd#(true(),s(x),s(y)) -> gcd#(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(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: gcd(|0|(),y) -> y
r7: gcd(s(x),|0|()) -> s(x)
r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),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(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: gcd(|0|(),y) -> y
r7: gcd(s(x),|0|()) -> s(x)
r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x))

The set of usable rules consists of

  r1, r2

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{3, x1 - 11}
            le_A(x1,x2) = max{7, x1 - 4, x2 - 4}
            |0|_A = 3
            true_A = 0
            false_A = 0
            le#!_A(x1,x2) = 0
            s!_A(x1) = 2
            le!_A(x1,x2) = max{3, x1 - 16, x2 - 21}
            |0|!_A = 2
            true!_A = 1
            false!_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{14, x1 + 6}
            le_A(x1,x2) = max{16, x1 - 15, x2 - 15}
            |0|_A = 14
            true_A = 0
            false_A = 0
            le#!_A(x1,x2) = max{1, x1 - 12, x2 - 15}
            s!_A(x1) = 2
            le!_A(x1,x2) = max{4, x1 - 15, x2 - 15}
            |0|!_A = 2
            true!_A = 3
            false!_A = 10
        
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []

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: gcd(|0|(),y) -> y
r7: gcd(s(x),|0|()) -> s(x)
r8: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r9: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r10: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x))

The set of usable rules consists of

  r1, r2

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) = 0
            s_A(x1) = max{3, x1 - 20}
            le_A(x1,x2) = max{16, x1 - 4, x2 - 4}
            |0|_A = 3
            true_A = 7
            false_A = 0
            minus#!_A(x1,x2) = 0
            s!_A(x1) = 7
            le!_A(x1,x2) = max{1, x1 - 31, x2 - 31}
            |0|!_A = 6
            true!_A = 5
            false!_A = 3
        
        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{7, x1 + 1}
            le_A(x1,x2) = max{2, x1 - 8, x2 - 1}
            |0|_A = 0
            true_A = 0
            false_A = 0
            minus#!_A(x1,x2) = max{4, x1 - 2, x2 - 8}
            s!_A(x1) = 3
            le!_A(x1,x2) = max{1, x1 - 8, x2 - 10}
            |0|!_A = 5
            true!_A = 4
            false!_A = 7
        
  
    partial status:
  
      pi(minus#) = []
      pi(s) = [1]
      pi(le) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.