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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_gcd#_A(x1,x2,x3) = max{5, x1 - 1, x2 + 3, x3 + 1}
          false_A = 9
          s_A(x1) = x1 + 17
          gcd#_A(x1,x2) = max{x1 + 3, x2 + 1}
          minus_A(x1,x2) = max{16, x1 + 14, x2 + 14}
          le_A(x1,x2) = max{21, x1 + 5, x2 + 5}
          true_A = 16
          |0|_A = 10
    
      precedence:
      
        true > if_gcd# = false = s = gcd# = minus = le = |0|
      
      partial status:
    
        pi(if_gcd#) = [3]
        pi(false) = []
        pi(s) = []
        pi(gcd#) = [2]
        pi(minus) = [1, 2]
        pi(le) = [1, 2]
        pi(true) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_gcd#_A(x1,x2,x3) = 7
          false_A = 16
          s_A(x1) = 7
          gcd#_A(x1,x2) = 7
          minus_A(x1,x2) = max{x1, x2 - 5}
          le_A(x1,x2) = 14
          true_A = 16
          |0|_A = 7
    
      precedence:
      
        if_gcd# = false = gcd# = |0| > s = true > minus = le
      
      partial status:
    
        pi(if_gcd#) = []
        pi(false) = []
        pi(s) = []
        pi(gcd#) = []
        pi(minus) = [1]
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- 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(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:

  {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(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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          gcd#_A(x1,x2) = x1 + 10
          s_A(x1) = max{24, x1 + 14}
          if_gcd#_A(x1,x2,x3) = max{6, x2 + 2}
          le_A(x1,x2) = max{x1 - 2, x2 + 25}
          true_A = 0
          minus_A(x1,x2) = max{16, x1 + 3}
          |0|_A = 14
          false_A = 13
    
      precedence:
      
        if_gcd# > gcd# > le = minus = |0| = false > true > s
      
      partial status:
    
        pi(gcd#) = [1]
        pi(s) = []
        pi(if_gcd#) = [2]
        pi(le) = [2]
        pi(true) = []
        pi(minus) = [1]
        pi(|0|) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          gcd#_A(x1,x2) = x1 + 8
          s_A(x1) = 0
          if_gcd#_A(x1,x2,x3) = 3
          le_A(x1,x2) = 9
          true_A = 10
          minus_A(x1,x2) = x1 + 5
          |0|_A = 0
          false_A = 10
    
      precedence:
      
        if_gcd# = false > gcd# = le = true = minus = |0| > s
      
      partial status:
    
        pi(gcd#) = []
        pi(s) = []
        pi(if_gcd#) = []
        pi(le) = []
        pi(true) = []
        pi(minus) = [1]
        pi(|0|) = []
        pi(false) = []
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  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(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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          le#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        minus# = s
      
      partial status:
    
        pi(minus#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        minus# = s
      
      partial status:
    
        pi(minus#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.