YES

We show the termination of the TRS R:

  minus(X,s(Y)) -> pred(minus(X,Y))
  minus(X,|0|()) -> X
  pred(s(X)) -> X
  le(s(X),s(Y)) -> le(X,Y)
  le(s(X),|0|()) -> false()
  le(|0|(),Y) -> true()
  gcd(|0|(),Y) -> |0|()
  gcd(s(X),|0|()) -> s(X)
  gcd(s(X),s(Y)) -> if(le(Y,X),s(X),s(Y))
  if(true(),s(X),s(Y)) -> gcd(minus(X,Y),s(Y))
  if(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: minus#(X,s(Y)) -> pred#(minus(X,Y))
p2: minus#(X,s(Y)) -> minus#(X,Y)
p3: le#(s(X),s(Y)) -> le#(X,Y)
p4: gcd#(s(X),s(Y)) -> if#(le(Y,X),s(X),s(Y))
p5: gcd#(s(X),s(Y)) -> le#(Y,X)
p6: if#(true(),s(X),s(Y)) -> gcd#(minus(X,Y),s(Y))
p7: if#(true(),s(X),s(Y)) -> minus#(X,Y)
p8: if#(false(),s(X),s(Y)) -> gcd#(minus(Y,X),s(X))
p9: if#(false(),s(X),s(Y)) -> minus#(Y,X)

and R consists of:

r1: minus(X,s(Y)) -> pred(minus(X,Y))
r2: minus(X,|0|()) -> X
r3: pred(s(X)) -> X
r4: le(s(X),s(Y)) -> le(X,Y)
r5: le(s(X),|0|()) -> false()
r6: le(|0|(),Y) -> true()
r7: gcd(|0|(),Y) -> |0|()
r8: gcd(s(X),|0|()) -> s(X)
r9: gcd(s(X),s(Y)) -> if(le(Y,X),s(X),s(Y))
r10: if(true(),s(X),s(Y)) -> gcd(minus(X,Y),s(Y))
r11: if(false(),s(X),s(Y)) -> gcd(minus(Y,X),s(X))

The estimated dependency graph contains the following SCCs:

  {p4, p6, p8}
  {p2}
  {p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: gcd#(s(X),s(Y)) -> if#(le(Y,X),s(X),s(Y))
p2: if#(false(),s(X),s(Y)) -> gcd#(minus(Y,X),s(X))
p3: if#(true(),s(X),s(Y)) -> gcd#(minus(X,Y),s(Y))

and R consists of:

r1: minus(X,s(Y)) -> pred(minus(X,Y))
r2: minus(X,|0|()) -> X
r3: pred(s(X)) -> X
r4: le(s(X),s(Y)) -> le(X,Y)
r5: le(s(X),|0|()) -> false()
r6: le(|0|(),Y) -> true()
r7: gcd(|0|(),Y) -> |0|()
r8: gcd(s(X),|0|()) -> s(X)
r9: gcd(s(X),s(Y)) -> if(le(Y,X),s(X),s(Y))
r10: if(true(),s(X),s(Y)) -> gcd(minus(X,Y),s(Y))
r11: if(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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      gcd#_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (1,0)
      s_A(x1) = x1 + (3,1)
      if#_A(x1,x2,x3) = x1 + x2 + x3
      le_A(x1,x2) = ((0,0),(1,0)) x1 + (1,1)
      false_A() = (1,1)
      minus_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + (1,1)
      true_A() = (0,3)
      pred_A(x1) = x1
      |0|_A() = (1,1)

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: minus#(X,s(Y)) -> minus#(X,Y)

and R consists of:

r1: minus(X,s(Y)) -> pred(minus(X,Y))
r2: minus(X,|0|()) -> X
r3: pred(s(X)) -> X
r4: le(s(X),s(Y)) -> le(X,Y)
r5: le(s(X),|0|()) -> false()
r6: le(|0|(),Y) -> true()
r7: gcd(|0|(),Y) -> |0|()
r8: gcd(s(X),|0|()) -> s(X)
r9: gcd(s(X),s(Y)) -> if(le(Y,X),s(X),s(Y))
r10: if(true(),s(X),s(Y)) -> gcd(minus(X,Y),s(Y))
r11: if(false(),s(X),s(Y)) -> gcd(minus(Y,X),s(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      minus#_A(x1,x2) = ((1,0),(1,0)) x2
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

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: le#(s(X),s(Y)) -> le#(X,Y)

and R consists of:

r1: minus(X,s(Y)) -> pred(minus(X,Y))
r2: minus(X,|0|()) -> X
r3: pred(s(X)) -> X
r4: le(s(X),s(Y)) -> le(X,Y)
r5: le(s(X),|0|()) -> false()
r6: le(|0|(),Y) -> true()
r7: gcd(|0|(),Y) -> |0|()
r8: gcd(s(X),|0|()) -> s(X)
r9: gcd(s(X),s(Y)) -> if(le(Y,X),s(X),s(Y))
r10: if(true(),s(X),s(Y)) -> gcd(minus(X,Y),s(Y))
r11: if(false(),s(X),s(Y)) -> gcd(minus(Y,X),s(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      le#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

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