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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2,x3,x4,x5) = ((1,1),(0,0)) x4 + ((1,1),(1,1)) x5 + (2,2)
          s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
          |0|_A() = (1,1)
  
    precedence:
    
      f# = |0| > s
    
    partial status:
  
      pi(f#) = [5]
      pi(s) = []
      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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2,x3,x4,x5) = ((0,1),(0,0)) x3 + ((1,1),(1,1)) x4 + x5 + (2,2)
          s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
          |0|_A() = (1,1)
  
    precedence:
    
      f# = s > |0|
    
    partial status:
  
      pi(f#) = []
      pi(s) = [1]
      pi(|0|) = []

The next rules are strictly ordered:

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2,x3,x4,x5) = ((0,1),(0,0)) x2 + x4 + (1,2)
          s_A(x1) = ((1,1),(1,1)) x1 + (2,2)
          |0|_A() = (0,1)
  
    precedence:
    
      f# = s = |0|
    
    partial status:
  
      pi(f#) = [4]
      pi(s) = [1]
      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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2,x3,x4,x5) = ((1,1),(1,1)) x2 + (2,2)
          s_A(x1) = ((1,1),(1,1)) x1 + (3,2)
          |0|_A() = (1,1)
  
    precedence:
    
      f# = s = |0|
    
    partial status:
  
      pi(f#) = []
      pi(s) = [1]
      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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2,x3,x4,x5) = ((0,1),(0,1)) x1 + x3 + x4 + x5 + (2,2)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      f# = s
    
    partial status:
  
      pi(f#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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