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. max/plus interpretations on natural numbers:
    
      if_gcd#_A(x1,x2,x3) = max{x1 + 3, x2, x3 - 2}
      false_A = 15
      s_A(x1) = x1 + 18
      gcd#_A(x1,x2) = max{2, x1 + 1, x2 - 2}
      minus_A(x1,x2) = x1 + 13
      le_A(x1,x2) = max{x1 + 13, x2 + 11}
      true_A = 10
      if_minus_A(x1,x2,x3) = x2 + 13
      |0|_A = 5
    
    2. max/plus interpretations on natural numbers:
    
      if_gcd#_A(x1,x2,x3) = 0
      false_A = 0
      s_A(x1) = max{1, x1}
      gcd#_A(x1,x2) = 0
      minus_A(x1,x2) = 1
      le_A(x1,x2) = 1
      true_A = 0
      if_minus_A(x1,x2,x3) = 1
      |0|_A = 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. max/plus interpretations on natural numbers:
    
      gcd#_A(x1,x2) = x1 + 1
      s_A(x1) = x1 + 5
      if_gcd#_A(x1,x2,x3) = max{x1, x2 + 1}
      le_A(x1,x2) = x2 + 6
      true_A = 1
      minus_A(x1,x2) = x1 + 4
      if_minus_A(x1,x2,x3) = x2 + 4
      |0|_A = 4
      false_A = 2
    
    2. max/plus interpretations on natural numbers:
    
      gcd#_A(x1,x2) = 1
      s_A(x1) = max{0, x1 - 2}
      if_gcd#_A(x1,x2,x3) = 0
      le_A(x1,x2) = 1
      true_A = 0
      minus_A(x1,x2) = 2
      if_minus_A(x1,x2,x3) = 1
      |0|_A = 0
      false_A = 0
    

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. max/plus interpretations on natural numbers:
    
      if_minus#_A(x1,x2,x3) = x2 + 1
      false_A = 1
      s_A(x1) = max{4, x1 + 3}
      minus#_A(x1,x2) = x1 + 3
      le_A(x1,x2) = max{x1 - 1, x2 + 2}
      |0|_A = 3
      true_A = 1
    
    2. max/plus interpretations on natural numbers:
    
      if_minus#_A(x1,x2,x3) = 1
      false_A = 0
      s_A(x1) = 0
      minus#_A(x1,x2) = 0
      le_A(x1,x2) = 1
      |0|_A = 0
      true_A = 0
    

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. max/plus interpretations on natural numbers:
    
      le#_A(x1,x2) = max{x1, x2}
      s_A(x1) = max{2, x1 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      le#_A(x1,x2) = 0
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p1

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