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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        if_gcd#_A(x1,x2,x3) = max{1, x1 - 12, x2 - 8, x3 - 9}
        false_A = 15
        s_A(x1) = x1 + 16
        gcd#_A(x1,x2) = max{0, x1 - 7, x2 - 9}
        minus_A(x1,x2) = max{x1 + 9, x2 + 13}
        le_A(x1,x2) = 16
        true_A = 15
        |0|_A = 11
  
    precedence:
    
      if_gcd# = false = s = gcd# = minus = le = true = |0|
    
    partial status:
  
      pi(if_gcd#) = []
      pi(false) = []
      pi(s) = [1]
      pi(gcd#) = []
      pi(minus) = [2]
      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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        gcd#_A(x1,x2) = x1 + 22
        s_A(x1) = max{10, x1 + 9}
        if_gcd#_A(x1,x2,x3) = x2 + 20
        le_A(x1,x2) = max{x1 + 32, x2 + 32}
        true_A = 0
        minus_A(x1,x2) = max{8, x1}
        |0|_A = 9
        false_A = 8
  
    precedence:
    
      |0| > s = if_gcd# = minus = false > gcd# = le = true
    
    partial status:
  
      pi(gcd#) = []
      pi(s) = [1]
      pi(if_gcd#) = []
      pi(le) = [2]
      pi(true) = []
      pi(minus) = [1]
      pi(|0|) = []
      pi(false) = []

The next rules are strictly ordered:

  p2

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))

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:

  (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(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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        le#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      le# = s
    
    partial status:
  
      pi(le#) = []
      pi(s) = []

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      minus# = s
    
    partial status:
  
      pi(minus#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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