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, 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_gcd#_A(x1,x2,x3) = max{2, x2 - 6, x3 - 3}
            false_A = 6
            s_A(x1) = 4
            gcd#_A(x1,x2) = max{2, x2 - 6}
            minus_A(x1,x2) = max{8, x1 - 3}
            le_A(x1,x2) = 5
            true_A = 1
            |0|_A = 1
            if_minus_A(x1,x2,x3) = x2
            if_gcd#!_A(x1,x2,x3) = max{x2 - 5, x3 + 9}
            false!_A = 16
            s!_A(x1) = max{11, x1 + 2}
            gcd#!_A(x1,x2) = max{13, x2 + 8}
            minus!_A(x1,x2) = max{x1 + 38, x2 + 13}
            le!_A(x1,x2) = max{x1 + 3, x2 + 14}
            true!_A = 3
            |0|!_A = 17
            if_minus!_A(x1,x2,x3) = max{x2 + 13, x3 + 12}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_gcd#_A(x1,x2,x3) = max{103, x2 + 14, x3 - 39}
            false_A = 7
            s_A(x1) = max{77, x1 + 65}
            gcd#_A(x1,x2) = max{103, x1 + 25, x2 - 39}
            minus_A(x1,x2) = max{2, x1}
            le_A(x1,x2) = max{6, x2 + 1}
            true_A = 1
            |0|_A = 41
            if_minus_A(x1,x2,x3) = max{66, x2}
            if_gcd#!_A(x1,x2,x3) = max{1, x2 - 67, x3 - 68}
            false!_A = 36
            s!_A(x1) = max{27, x1 + 25}
            gcd#!_A(x1,x2) = max{2, x1 - 66, x2 - 68}
            minus!_A(x1,x2) = max{x1 - 4, x2 + 29}
            le!_A(x1,x2) = max{75, x1 - 2}
            true!_A = 34
            |0|!_A = 37
            if_minus!_A(x1,x2,x3) = max{37, x2 - 53, x3 + 28}
        
  
    partial status:
  
      pi(if_gcd#) = []
      pi(false) = []
      pi(s) = []
      pi(gcd#) = []
      pi(minus) = [1]
      pi(le) = [1, 2]
      pi(true) = []
      pi(|0|) = []
      pi(if_minus) = [2]

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),s(y))
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),s(y))
p2: 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:
  
      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:
          
            gcd#_A(x1,x2) = 1
            s_A(x1) = x1 + 20
            if_gcd#_A(x1,x2,x3) = 1
            le_A(x1,x2) = max{1, x1 - 21}
            true_A = 2
            minus_A(x1,x2) = x1 + 19
            |0|_A = 39
            false_A = 1
            if_minus_A(x1,x2,x3) = max{38, x2 + 19}
            gcd#!_A(x1,x2) = max{11, x1 - 4}
            s!_A(x1) = 12
            if_gcd#!_A(x1,x2,x3) = max{8, x2 - 5}
            le!_A(x1,x2) = 16
            true!_A = 10
            minus!_A(x1,x2) = max{9, x1 - 16}
            |0|!_A = 7
            false!_A = 0
            if_minus!_A(x1,x2,x3) = max{8, x1 + 4, x2 - 17}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gcd#_A(x1,x2) = 15
            s_A(x1) = 19
            if_gcd#_A(x1,x2,x3) = max{15, x3 - 5}
            le_A(x1,x2) = 9
            true_A = 1
            minus_A(x1,x2) = 28
            |0|_A = 1
            false_A = 2
            if_minus_A(x1,x2,x3) = max{19, x1 - 16, x2 - 20}
            gcd#!_A(x1,x2) = max{40, x1 - 28}
            s!_A(x1) = max{40, x1 + 8}
            if_gcd#!_A(x1,x2,x3) = 41
            le!_A(x1,x2) = 39
            true!_A = 43
            minus!_A(x1,x2) = 42
            |0|!_A = 7
            false!_A = 37
            if_minus!_A(x1,x2,x3) = max{6, x1 - 2, x2 - 5}
        
  
    partial status:
  
      pi(gcd#) = []
      pi(s) = []
      pi(if_gcd#) = []
      pi(le) = []
      pi(true) = []
      pi(minus) = [1]
      pi(|0|) = []
      pi(false) = []
      pi(if_minus) = [2]

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: 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:
  
      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 - 21, x2 - 37, x3 - 23}
            false_A = 0
            s_A(x1) = max{35, x1 + 7}
            minus#_A(x1,x2) = max{0, x1 - 37, x2 - 23}
            le_A(x1,x2) = max{20, x1 - 36, x2 - 2}
            |0|_A = 1
            true_A = 0
            minus_A(x1,x2) = max{71, x1 - 37, x2 - 24}
            if_minus#!_A(x1,x2,x3) = max{5, x1 - 74, x2 - 66, x3 - 53}
            false!_A = 2
            s!_A(x1) = max{6, x1 - 103}
            minus#!_A(x1,x2) = max{5, x1 - 60, x2 - 53}
            le!_A(x1,x2) = max{4, x1 - 87, x2 - 104}
            |0|!_A = 1
            true!_A = 2
            minus!_A(x1,x2) = max{3, x1 - 58, x2 - 51}
        
        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 - 55, x3 - 22}
            false_A = 0
            s_A(x1) = max{34, x1 + 4}
            minus#_A(x1,x2) = max{0, x1 - 55, x2 - 22}
            le_A(x1,x2) = max{14, x1 - 40, x2 - 7}
            |0|_A = 1
            true_A = 0
            minus_A(x1,x2) = max{36, x1 - 36, x2 - 23}
            if_minus#!_A(x1,x2,x3) = max{6, x1 - 67, x2 - 105, x3 - 72}
            false!_A = 2
            s!_A(x1) = max{5, x1 - 63}
            minus#!_A(x1,x2) = max{6, x1 - 101, x2 - 72}
            le!_A(x1,x2) = max{5, x1 - 100, x2 - 35}
            |0|!_A = 1
            true!_A = 4
            minus!_A(x1,x2) = max{3, x1 - 65, x2 - 65}
        
  
    partial status:
  
      pi(if_minus#) = [2]
      pi(false) = []
      pi(s) = [1]
      pi(minus#) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = [1]

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: 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:

  (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: 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, 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!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{2, x1 - 20}
            le_A(x1,x2) = max{2, x1 - 10, x2 - 6}
            |0|_A = 5
            true_A = 0
            minus_A(x1,x2) = max{19, x1 - 16, x2 - 16}
            le#!_A(x1,x2) = 0
            s!_A(x1) = 16
            le!_A(x1,x2) = max{18, x1 - 24, x2 - 28}
            |0|!_A = 14
            true!_A = 15
            minus!_A(x1,x2) = max{17, x1 - 25, x2 - 29}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = max{9, x1 - 20}
            le_A(x1,x2) = max{2, x1 - 12, x2 - 12}
            |0|_A = 11
            true_A = 0
            minus_A(x1,x2) = max{19, x1 - 18, x2 - 18}
            le#!_A(x1,x2) = max{1, x1 - 10, x2 - 73}
            s!_A(x1) = max{4, x1 - 73}
            le!_A(x1,x2) = max{6, x1 - 12, x2 - 10}
            |0|!_A = 3
            true!_A = 7
            minus!_A(x1,x2) = max{5, x1 - 32, x2 - 30}
        
  
    partial status:
  
      pi(le#) = [1]
      pi(s) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(minus) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.