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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_gcd#_A(x1,x2,x3) = max{21, x2 + 5, x3 + 4}
          false_A = 1
          s_A(x1) = x1 + 14
          gcd#_A(x1,x2) = max{20, x1 + 7, x2 + 4}
          minus_A(x1,x2) = x1 + 4
          le_A(x1,x2) = max{7, x1 - 6}
          true_A = 0
          if_minus_A(x1,x2,x3) = max{x1 + 9, x2 + 4}
          |0|_A = 3
    
      precedence:
      
        minus > if_minus > s > le = true > false = |0| > if_gcd# = gcd#
      
      partial status:
    
        pi(if_gcd#) = [3]
        pi(false) = []
        pi(s) = [1]
        pi(gcd#) = [2]
        pi(minus) = []
        pi(le) = []
        pi(true) = []
        pi(if_minus) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_gcd#_A(x1,x2,x3) = x3 + 13
          false_A = 1
          s_A(x1) = max{13, x1 - 2}
          gcd#_A(x1,x2) = x2 + 13
          minus_A(x1,x2) = 25
          le_A(x1,x2) = 2
          true_A = 27
          if_minus_A(x1,x2,x3) = max{24, x1 + 1}
          |0|_A = 26
    
      precedence:
      
        minus = le = true > if_gcd# = s = gcd# = if_minus = |0| > false
      
      partial status:
    
        pi(if_gcd#) = []
        pi(false) = []
        pi(s) = []
        pi(gcd#) = []
        pi(minus) = []
        pi(le) = []
        pi(true) = []
        pi(if_minus) = [1]
        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(|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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          gcd#_A(x1,x2) = x1 + 20
          s_A(x1) = max{31, x1 + 27}
          if_gcd#_A(x1,x2,x3) = max{30, x1 + 1, x2 + 1}
          le_A(x1,x2) = max{45, x2 + 19}
          true_A = 7
          minus_A(x1,x2) = x1 + 7
          if_minus_A(x1,x2,x3) = max{38, x2 + 7}
          |0|_A = 0
          false_A = 0
    
      precedence:
      
        le = minus > s = true = if_minus = |0| = false > gcd# > if_gcd#
      
      partial status:
    
        pi(gcd#) = [1]
        pi(s) = []
        pi(if_gcd#) = [1, 2]
        pi(le) = [2]
        pi(true) = []
        pi(minus) = [1]
        pi(if_minus) = [2]
        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) = 5
          if_gcd#_A(x1,x2,x3) = max{4, x1, x2 + 3}
          le_A(x1,x2) = 13
          true_A = 15
          minus_A(x1,x2) = 6
          if_minus_A(x1,x2,x3) = max{7, x2}
          |0|_A = 2
          false_A = 3
    
      precedence:
      
        gcd# = s = if_gcd# = le = true = minus = if_minus = |0| = false
      
      partial status:
    
        pi(gcd#) = [1]
        pi(s) = []
        pi(if_gcd#) = []
        pi(le) = []
        pi(true) = []
        pi(minus) = []
        pi(if_minus) = [2]
        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: 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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_minus#_A(x1,x2,x3) = max{x1 + 2, x2 + 1, x3 + 4}
          false_A = 4
          s_A(x1) = x1 + 5
          minus#_A(x1,x2) = max{x1 + 6, x2 + 4}
          le_A(x1,x2) = max{x1 + 3, x2 - 6}
          |0|_A = 11
          true_A = 12
    
      precedence:
      
        if_minus# = s = minus# = le = |0| > false = true
      
      partial status:
    
        pi(if_minus#) = [2]
        pi(false) = []
        pi(s) = [1]
        pi(minus#) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_minus#_A(x1,x2,x3) = max{1, x2}
          false_A = 7
          s_A(x1) = x1 + 8
          minus#_A(x1,x2) = x1
          le_A(x1,x2) = 1
          |0|_A = 0
          true_A = 2
    
      precedence:
      
        true > if_minus# = false = s = minus# = le = |0|
      
      partial status:
    
        pi(if_minus#) = []
        pi(false) = []
        pi(s) = [1]
        pi(minus#) = [1]
        pi(le) = []
        pi(|0|) = []
        pi(true) = []
    

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(|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

  (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.