YES

We show the termination of the TRS R:

  f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5)
  f(|0|(),s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5)
  f(|0|(),|0|(),s(x3),x4,x5) -> f(x3,x3,x3,x4,x5)
  f(|0|(),|0|(),|0|(),s(x4),x5) -> f(x4,x4,x4,x4,x5)
  f(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f(x5,x5,x5,x5,x5)
  f(|0|(),|0|(),|0|(),|0|(),|0|()) -> |0|()

-- SCC decomposition.

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

p1: f#(s(x1),x2,x3,x4,x5) -> f#(x1,x2,x3,x4,x5)
p2: f#(|0|(),s(x2),x3,x4,x5) -> f#(x2,x2,x3,x4,x5)
p3: f#(|0|(),|0|(),s(x3),x4,x5) -> f#(x3,x3,x3,x4,x5)
p4: f#(|0|(),|0|(),|0|(),s(x4),x5) -> f#(x4,x4,x4,x4,x5)
p5: f#(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f#(x5,x5,x5,x5,x5)

and R consists of:

r1: f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5)
r2: f(|0|(),s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5)
r3: f(|0|(),|0|(),s(x3),x4,x5) -> f(x3,x3,x3,x4,x5)
r4: f(|0|(),|0|(),|0|(),s(x4),x5) -> f(x4,x4,x4,x4,x5)
r5: f(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f(x5,x5,x5,x5,x5)
r6: f(|0|(),|0|(),|0|(),|0|(),|0|()) -> |0|()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: f#(s(x1),x2,x3,x4,x5) -> f#(x1,x2,x3,x4,x5)
p2: f#(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f#(x5,x5,x5,x5,x5)
p3: f#(|0|(),|0|(),|0|(),s(x4),x5) -> f#(x4,x4,x4,x4,x5)
p4: f#(|0|(),|0|(),s(x3),x4,x5) -> f#(x3,x3,x3,x4,x5)
p5: f#(|0|(),s(x2),x3,x4,x5) -> f#(x2,x2,x3,x4,x5)

and R consists of:

r1: f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5)
r2: f(|0|(),s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5)
r3: f(|0|(),|0|(),s(x3),x4,x5) -> f(x3,x3,x3,x4,x5)
r4: f(|0|(),|0|(),|0|(),s(x4),x5) -> f(x4,x4,x4,x4,x5)
r5: f(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f(x5,x5,x5,x5,x5)
r6: f(|0|(),|0|(),|0|(),|0|(),|0|()) -> |0|()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1,x2,x3,x4,x5) = ((0,0),(1,0)) x5
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      |0|_A() = (1,1)

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: f#(s(x1),x2,x3,x4,x5) -> f#(x1,x2,x3,x4,x5)
p2: f#(|0|(),|0|(),|0|(),s(x4),x5) -> f#(x4,x4,x4,x4,x5)
p3: f#(|0|(),|0|(),s(x3),x4,x5) -> f#(x3,x3,x3,x4,x5)
p4: f#(|0|(),s(x2),x3,x4,x5) -> f#(x2,x2,x3,x4,x5)

and R consists of:

r1: f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5)
r2: f(|0|(),s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5)
r3: f(|0|(),|0|(),s(x3),x4,x5) -> f(x3,x3,x3,x4,x5)
r4: f(|0|(),|0|(),|0|(),s(x4),x5) -> f(x4,x4,x4,x4,x5)
r5: f(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f(x5,x5,x5,x5,x5)
r6: f(|0|(),|0|(),|0|(),|0|(),|0|()) -> |0|()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: f#(s(x1),x2,x3,x4,x5) -> f#(x1,x2,x3,x4,x5)
p2: f#(|0|(),s(x2),x3,x4,x5) -> f#(x2,x2,x3,x4,x5)
p3: f#(|0|(),|0|(),s(x3),x4,x5) -> f#(x3,x3,x3,x4,x5)
p4: f#(|0|(),|0|(),|0|(),s(x4),x5) -> f#(x4,x4,x4,x4,x5)

and R consists of:

r1: f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5)
r2: f(|0|(),s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5)
r3: f(|0|(),|0|(),s(x3),x4,x5) -> f(x3,x3,x3,x4,x5)
r4: f(|0|(),|0|(),|0|(),s(x4),x5) -> f(x4,x4,x4,x4,x5)
r5: f(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f(x5,x5,x5,x5,x5)
r6: f(|0|(),|0|(),|0|(),|0|(),|0|()) -> |0|()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1,x2,x3,x4,x5) = ((1,0),(1,1)) x4
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      |0|_A() = (1,1)

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: f#(s(x1),x2,x3,x4,x5) -> f#(x1,x2,x3,x4,x5)
p2: f#(|0|(),s(x2),x3,x4,x5) -> f#(x2,x2,x3,x4,x5)
p3: f#(|0|(),|0|(),s(x3),x4,x5) -> f#(x3,x3,x3,x4,x5)

and R consists of:

r1: f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5)
r2: f(|0|(),s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5)
r3: f(|0|(),|0|(),s(x3),x4,x5) -> f(x3,x3,x3,x4,x5)
r4: f(|0|(),|0|(),|0|(),s(x4),x5) -> f(x4,x4,x4,x4,x5)
r5: f(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f(x5,x5,x5,x5,x5)
r6: f(|0|(),|0|(),|0|(),|0|(),|0|()) -> |0|()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: f#(s(x1),x2,x3,x4,x5) -> f#(x1,x2,x3,x4,x5)
p2: f#(|0|(),|0|(),s(x3),x4,x5) -> f#(x3,x3,x3,x4,x5)
p3: f#(|0|(),s(x2),x3,x4,x5) -> f#(x2,x2,x3,x4,x5)

and R consists of:

r1: f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5)
r2: f(|0|(),s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5)
r3: f(|0|(),|0|(),s(x3),x4,x5) -> f(x3,x3,x3,x4,x5)
r4: f(|0|(),|0|(),|0|(),s(x4),x5) -> f(x4,x4,x4,x4,x5)
r5: f(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f(x5,x5,x5,x5,x5)
r6: f(|0|(),|0|(),|0|(),|0|(),|0|()) -> |0|()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1,x2,x3,x4,x5) = ((0,0),(1,0)) x2 + x3
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      |0|_A() = (1,1)

The next rules are strictly ordered:

  p2, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: f#(s(x1),x2,x3,x4,x5) -> f#(x1,x2,x3,x4,x5)

and R consists of:

r1: f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5)
r2: f(|0|(),s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5)
r3: f(|0|(),|0|(),s(x3),x4,x5) -> f(x3,x3,x3,x4,x5)
r4: f(|0|(),|0|(),|0|(),s(x4),x5) -> f(x4,x4,x4,x4,x5)
r5: f(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f(x5,x5,x5,x5,x5)
r6: f(|0|(),|0|(),|0|(),|0|(),|0|()) -> |0|()

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(s(x1),x2,x3,x4,x5) -> f#(x1,x2,x3,x4,x5)

and R consists of:

r1: f(s(x1),x2,x3,x4,x5) -> f(x1,x2,x3,x4,x5)
r2: f(|0|(),s(x2),x3,x4,x5) -> f(x2,x2,x3,x4,x5)
r3: f(|0|(),|0|(),s(x3),x4,x5) -> f(x3,x3,x3,x4,x5)
r4: f(|0|(),|0|(),|0|(),s(x4),x5) -> f(x4,x4,x4,x4,x5)
r5: f(|0|(),|0|(),|0|(),|0|(),s(x5)) -> f(x5,x5,x5,x5,x5)
r6: f(|0|(),|0|(),|0|(),|0|(),|0|()) -> |0|()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1,x2,x3,x4,x5) = ((1,0),(1,1)) x1
      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.