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)
  pred(s(x)) -> x
  minus(x,|0|()) -> x
  minus(x,s(y)) -> pred(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#(x,s(y)) -> pred#(minus(x,y))
p3: minus#(x,s(y)) -> minus#(x,y)
p4: gcd#(s(x),s(y)) -> if_gcd#(le(y,x),s(x),s(y))
p5: gcd#(s(x),s(y)) -> le#(y,x)
p6: if_gcd#(true(),s(x),s(y)) -> gcd#(minus(x,y),s(y))
p7: if_gcd#(true(),s(x),s(y)) -> minus#(x,y)
p8: if_gcd#(false(),s(x),s(y)) -> gcd#(minus(y,x),s(x))
p9: 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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: gcd(|0|(),y) -> y
r8: gcd(s(x),|0|()) -> s(x)
r9: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r10: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r11: if_gcd(false(),s(x),s(y)) -> gcd(minus(y,x),s(x))

The estimated dependency graph contains the following SCCs:

  {p4, p6, p8}
  {p1}
  {p3}


-- 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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: gcd(|0|(),y) -> y
r8: gcd(s(x),|0|()) -> s(x)
r9: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r10: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r11: 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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        if_gcd#_A(x1,x2,x3) = x2 + x3
        false_A() = 1
        s_A(x1) = x1 + 3
        gcd#_A(x1,x2) = x1 + x2 + 1
        minus_A(x1,x2) = x1 + 1
        le_A(x1,x2) = x1 + x2 + 1
        true_A() = 1
        pred_A(x1) = x1
        |0|_A() = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        true > false > |0| > pred > s > minus > le > if_gcd# > gcd#
      
      argument filter:
    
        pi(if_gcd#) = 3
        pi(false) = []
        pi(s) = 1
        pi(gcd#) = [1, 2]
        pi(minus) = 1
        pi(le) = []
        pi(true) = []
        pi(pred) = 1
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: gcd(|0|(),y) -> y
r8: gcd(s(x),|0|()) -> s(x)
r9: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r10: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r11: 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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        le#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > le#
      
      argument filter:
    
        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#(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: pred(s(x)) -> x
r5: minus(x,|0|()) -> x
r6: minus(x,s(y)) -> pred(minus(x,y))
r7: gcd(|0|(),y) -> y
r8: gcd(s(x),|0|()) -> s(x)
r9: gcd(s(x),s(y)) -> if_gcd(le(y,x),s(x),s(y))
r10: if_gcd(true(),s(x),s(y)) -> gcd(minus(x,y),s(y))
r11: 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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        minus#_A(x1,x2) = x2
        s_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > minus#
      
      argument filter:
    
        pi(minus#) = 2
        pi(s) = 1
    

The next rules are strictly ordered:

  p1

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