YES

We show the termination of the TRS R:

  min(x,|0|()) -> |0|()
  min(|0|(),y) -> |0|()
  min(s(x),s(y)) -> s(min(x,y))
  max(x,|0|()) -> x
  max(|0|(),y) -> y
  max(s(x),s(y)) -> s(max(x,y))
  -(x,|0|()) -> x
  -(s(x),s(y)) -> -(x,y)
  gcd(s(x),|0|()) -> s(x)
  gcd(|0|(),s(x)) -> s(x)
  gcd(s(x),s(y)) -> gcd(-(max(x,y),min(x,y)),s(min(x,y)))

-- SCC decomposition.

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

p1: min#(s(x),s(y)) -> min#(x,y)
p2: max#(s(x),s(y)) -> max#(x,y)
p3: -#(s(x),s(y)) -> -#(x,y)
p4: gcd#(s(x),s(y)) -> gcd#(-(max(x,y),min(x,y)),s(min(x,y)))
p5: gcd#(s(x),s(y)) -> -#(max(x,y),min(x,y))
p6: gcd#(s(x),s(y)) -> max#(x,y)
p7: gcd#(s(x),s(y)) -> min#(x,y)

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: -(x,|0|()) -> x
r8: -(s(x),s(y)) -> -(x,y)
r9: gcd(s(x),|0|()) -> s(x)
r10: gcd(|0|(),s(x)) -> s(x)
r11: gcd(s(x),s(y)) -> gcd(-(max(x,y),min(x,y)),s(min(x,y)))

The estimated dependency graph contains the following SCCs:

  {p4}
  {p1}
  {p2}
  {p3}


-- Reduction pair.

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

p1: gcd#(s(x),s(y)) -> gcd#(-(max(x,y),min(x,y)),s(min(x,y)))

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: -(x,|0|()) -> x
r8: -(s(x),s(y)) -> -(x,y)
r9: gcd(s(x),|0|()) -> s(x)
r10: gcd(|0|(),s(x)) -> s(x)
r11: gcd(s(x),s(y)) -> gcd(-(max(x,y),min(x,y)),s(min(x,y)))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      gcd#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,0)) x2
      s_A(x1) = ((1,1),(1,1)) x1 + (0,1)
      -_A(x1,x2) = x1
      max_A(x1,x2) = x1 + x2
      min_A(x1,x2) = x1
      |0|_A() = (0,0)

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: min#(s(x),s(y)) -> min#(x,y)

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: -(x,|0|()) -> x
r8: -(s(x),s(y)) -> -(x,y)
r9: gcd(s(x),|0|()) -> s(x)
r10: gcd(|0|(),s(x)) -> s(x)
r11: gcd(s(x),s(y)) -> gcd(-(max(x,y),min(x,y)),s(min(x,y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

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: max#(s(x),s(y)) -> max#(x,y)

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: -(x,|0|()) -> x
r8: -(s(x),s(y)) -> -(x,y)
r9: gcd(s(x),|0|()) -> s(x)
r10: gcd(|0|(),s(x)) -> s(x)
r11: gcd(s(x),s(y)) -> gcd(-(max(x,y),min(x,y)),s(min(x,y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

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: -#(s(x),s(y)) -> -#(x,y)

and R consists of:

r1: min(x,|0|()) -> |0|()
r2: min(|0|(),y) -> |0|()
r3: min(s(x),s(y)) -> s(min(x,y))
r4: max(x,|0|()) -> x
r5: max(|0|(),y) -> y
r6: max(s(x),s(y)) -> s(max(x,y))
r7: -(x,|0|()) -> x
r8: -(s(x),s(y)) -> -(x,y)
r9: gcd(s(x),|0|()) -> s(x)
r10: gcd(|0|(),s(x)) -> s(x)
r11: gcd(s(x),s(y)) -> gcd(-(max(x,y),min(x,y)),s(min(x,y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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