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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > f# > s
      
      argument filter:
    
        pi(f#) = [5]
        pi(s) = 1
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2,x3,x4,x5) = x5
        s_A(x1) = x1 + 1
        |0|_A() = 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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        f# > |0| > s
      
      argument filter:
    
        pi(f#) = [4]
        pi(s) = 1
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2,x3,x4,x5) = x4
        s_A(x1) = x1 + 1
        |0|_A() = 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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > f# > |0|
      
      argument filter:
    
        pi(f#) = [3]
        pi(s) = 1
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2,x3,x4,x5) = x3
        s_A(x1) = x1 + 1
        |0|_A() = 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|(),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}


-- 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)

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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > s > f#
      
      argument filter:
    
        pi(f#) = [1, 2]
        pi(s) = [1]
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2,x3,x4,x5) = x1 + x2
        s_A(x1) = x1 + 1
        |0|_A() = 1
    

The next rules are strictly ordered:

  p1, p2

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