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))
  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),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: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),s(y))
p6: gcd#(s(x),s(y)) -> le#(y,x)
p7: if_gcd#(true(),s(x),s(y)) -> gcd#(minus(x,y),s(y))
p8: if_gcd#(true(),s(x),s(y)) -> minus#(x,y)
p9: if_gcd#(false(),s(x),s(y)) -> gcd#(minus(y,x),s(x))
p10: 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(|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: gcd(|0|(),y) -> y
r9: gcd(s(x),|0|()) -> s(x)
r10: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r11: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r12: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x))

The estimated dependency graph contains the following SCCs:

  {p5, p7, p9}
  {p2, p4}
  {p1}


-- 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(|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: gcd(|0|(),y) -> y
r9: gcd(s(x),|0|()) -> s(x)
r10: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r11: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r12: 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, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_gcd#_A(x1,x2,x3) = max{37, x1 + 2, x2 - 21, x3 - 40}
        false_A = 2
        s_A(x1) = max{49, x1 + 29}
        gcd#_A(x1,x2) = max{19, x1 - 12, x2 - 40}
        minus_A(x1,x2) = max{19, x1}
        le_A(x1,x2) = max{18, x2 - 48}
        true_A = 4
        |0|_A = 49
        if_minus_A(x1,x2,x3) = max{28, x2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            if_gcd#_A(x1,x2,x3) = x2 + x3
            false_A() = 1
            s_A(x1) = x1 + 4
            gcd#_A(x1,x2) = x1 + x2
            minus_A(x1,x2) = x1 + 1
            le_A(x1,x2) = x1 + 1
            true_A() = 1
            |0|_A() = 3
            if_minus_A(x1,x2,x3) = x2 + 1
            if_gcd#!_A(x1,x2,x3) = x2 + 1
            false!_A() = 6
            s!_A(x1) = 8
            gcd#!_A(x1,x2) = x1 + 3
            minus!_A(x1,x2) = x1 + 5
            le!_A(x1,x2) = x1 + x2 + 9
            true!_A() = 3
            |0|!_A() = 7
            if_minus!_A(x1,x2,x3) = 8
  
    partial status:
  
      pi(if_gcd#) = [1]
      pi(false) = []
      pi(s) = []
      pi(gcd#) = []
      pi(minus) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(if_minus) = []

The following rules are removed:

  p1, p2, p3

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: gcd(|0|(),y) -> y
r9: gcd(s(x),|0|()) -> s(x)
r10: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r11: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r12: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),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:
  
      max/plus interpretation on natural numbers:
      
        if_minus#_A(x1,x2,x3) = max{12, x1 - 10, x2 + 1}
        false_A = 5
        s_A(x1) = max{11, x1 + 4}
        minus#_A(x1,x2) = max{6, x1 + 1}
        le_A(x1,x2) = max{1, x1 - 2}
        |0|_A = 10
        true_A = 7
        minus_A(x1,x2) = max{17, x1 + 10, x2 - 1}
  
    base order 2:
  
      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) = 0
            false_A() = 1
            s_A(x1) = 3
            minus#_A(x1,x2) = 0
            le_A(x1,x2) = 1
            |0|_A() = 3
            true_A() = 1
            minus_A(x1,x2) = x1 + x2 + 5
            if_minus#!_A(x1,x2,x3) = 5
            false!_A() = 10
            s!_A(x1) = x1 + 4
            minus#!_A(x1,x2) = x1 + 3
            le!_A(x1,x2) = 1
            |0|!_A() = 3
            true!_A() = 0
            minus!_A(x1,x2) = 11
  
    partial status:
  
      pi(if_minus#) = []
      pi(false) = []
      pi(s) = [1]
      pi(minus#) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = [1]

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: gcd(|0|(),y) -> y
r9: gcd(s(x),|0|()) -> s(x)
r10: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r11: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r12: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x))

The set of usable rules consists of

  r4

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = max{12, x2 - 1}
        s_A(x1) = x1 + 16
        minus_A(x1,x2) = max{10, x1 - 16}
        |0|_A = 9
  
    base order 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 + 2
            minus_A(x1,x2) = x1 + x2 + 4
            |0|_A() = 1
            le#!_A(x1,x2) = 0
            s!_A(x1) = x1 + 4
            minus!_A(x1,x2) = 1
            |0|!_A() = 8
  
    partial status:
  
      pi(le#) = []
      pi(s) = [1]
      pi(minus) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.