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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        gcd#_A(x1,x2) = max{2, x1, x2 - 4}
        s_A(x1) = x1 + 7
        if#_A(x1,x2,x3) = max{0, x1 - 2, x2 - 1, x3 - 4}
        le_A(x1,x2) = x1 + 5
        false_A = 6
        minus_A(x1,x2) = max{x1 + 2, x2 + 3}
        true_A = 2
        pred_A(x1) = max{10, x1}
        |0|_A = 7
  
    precedence:
    
      false > gcd# = if# = le = minus = pred = |0| > s = true
    
    partial status:
  
      pi(gcd#) = []
      pi(s) = [1]
      pi(if#) = []
      pi(le) = [1]
      pi(false) = []
      pi(minus) = [1, 2]
      pi(true) = []
      pi(pred) = []
      pi(|0|) = []

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#(le(Y,X),s(X),s(Y))
p2: 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 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#(le(Y,X),s(X),s(Y))
p2: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        gcd#_A(x1,x2) = max{0, x1 - 5}
        s_A(x1) = x1 + 15
        if#_A(x1,x2,x3) = max{1, x2 - 9}
        le_A(x1,x2) = max{x1 + 14, x2 + 14}
        true_A = 13
        minus_A(x1,x2) = x1 + 11
        pred_A(x1) = max{11, x1}
        |0|_A = 12
        false_A = 13
  
    precedence:
    
      gcd# = s = if# = le = true = minus = pred = |0| = false
    
    partial status:
  
      pi(gcd#) = []
      pi(s) = [1]
      pi(if#) = []
      pi(le) = [2]
      pi(true) = []
      pi(minus) = []
      pi(pred) = []
      pi(|0|) = []
      pi(false) = []

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: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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 monotone reduction pair:

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