YES

We show the termination of the TRS R:

  not(not(x)) -> x
  not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
  not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

-- SCC decomposition.

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

p1: not#(or(x,y)) -> not#(not(not(x)))
p2: not#(or(x,y)) -> not#(not(x))
p3: not#(or(x,y)) -> not#(x)
p4: not#(or(x,y)) -> not#(not(not(y)))
p5: not#(or(x,y)) -> not#(not(y))
p6: not#(or(x,y)) -> not#(y)
p7: not#(and(x,y)) -> not#(not(not(x)))
p8: not#(and(x,y)) -> not#(not(x))
p9: not#(and(x,y)) -> not#(x)
p10: not#(and(x,y)) -> not#(not(not(y)))
p11: not#(and(x,y)) -> not#(not(y))
p12: not#(and(x,y)) -> not#(y)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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

p1: not#(or(x,y)) -> not#(not(not(x)))
p2: not#(and(x,y)) -> not#(y)
p3: not#(and(x,y)) -> not#(not(y))
p4: not#(and(x,y)) -> not#(not(not(y)))
p5: not#(and(x,y)) -> not#(x)
p6: not#(and(x,y)) -> not#(not(x))
p7: not#(and(x,y)) -> not#(not(not(x)))
p8: not#(or(x,y)) -> not#(y)
p9: not#(or(x,y)) -> not#(not(y))
p10: not#(or(x,y)) -> not#(not(not(y)))
p11: not#(or(x,y)) -> not#(x)
p12: not#(or(x,y)) -> not#(not(x))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = ((1,0),(1,1)) x1 + (1,6)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (4,1)
          not_A(x1) = ((1,0),(1,1)) x1 + (0,2)
          and_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (4,5)
  
    precedence:
    
      not# = or > not = and
    
    partial status:
  
      pi(not#) = [1]
      pi(or) = []
      pi(not) = [1]
      pi(and) = []

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: not#(or(x,y)) -> not#(not(not(x)))
p2: not#(and(x,y)) -> not#(y)
p3: not#(and(x,y)) -> not#(not(y))
p4: not#(and(x,y)) -> not#(x)
p5: not#(and(x,y)) -> not#(not(x))
p6: not#(and(x,y)) -> not#(not(not(x)))
p7: not#(or(x,y)) -> not#(y)
p8: not#(or(x,y)) -> not#(not(y))
p9: not#(or(x,y)) -> not#(not(not(y)))
p10: not#(or(x,y)) -> not#(x)
p11: not#(or(x,y)) -> not#(not(x))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

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

p1: not#(or(x,y)) -> not#(not(not(x)))
p2: not#(or(x,y)) -> not#(not(x))
p3: not#(or(x,y)) -> not#(x)
p4: not#(or(x,y)) -> not#(not(not(y)))
p5: not#(or(x,y)) -> not#(not(y))
p6: not#(or(x,y)) -> not#(y)
p7: not#(and(x,y)) -> not#(not(not(x)))
p8: not#(and(x,y)) -> not#(not(x))
p9: not#(and(x,y)) -> not#(x)
p10: not#(and(x,y)) -> not#(not(y))
p11: not#(and(x,y)) -> not#(y)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = x1 + (1,2)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,1)
          not_A(x1) = ((1,0),(1,1)) x1
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,1)
  
    precedence:
    
      not# = or = and > not
    
    partial status:
  
      pi(not#) = [1]
      pi(or) = []
      pi(not) = [1]
      pi(and) = []

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: not#(or(x,y)) -> not#(not(x))
p2: not#(or(x,y)) -> not#(x)
p3: not#(or(x,y)) -> not#(not(not(y)))
p4: not#(or(x,y)) -> not#(not(y))
p5: not#(or(x,y)) -> not#(y)
p6: not#(and(x,y)) -> not#(not(not(x)))
p7: not#(and(x,y)) -> not#(not(x))
p8: not#(and(x,y)) -> not#(x)
p9: not#(and(x,y)) -> not#(not(y))
p10: not#(and(x,y)) -> not#(y)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

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

p1: not#(or(x,y)) -> not#(not(x))
p2: not#(and(x,y)) -> not#(y)
p3: not#(and(x,y)) -> not#(not(y))
p4: not#(and(x,y)) -> not#(x)
p5: not#(and(x,y)) -> not#(not(x))
p6: not#(and(x,y)) -> not#(not(not(x)))
p7: not#(or(x,y)) -> not#(y)
p8: not#(or(x,y)) -> not#(not(y))
p9: not#(or(x,y)) -> not#(not(not(y)))
p10: not#(or(x,y)) -> not#(x)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = ((0,0),(1,0)) x1 + (1,4)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,5)
          not_A(x1) = ((1,0),(1,1)) x1 + (0,2)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,5)
  
    precedence:
    
      not# = or = and > not
    
    partial status:
  
      pi(not#) = []
      pi(or) = []
      pi(not) = [1]
      pi(and) = []

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

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

p1: not#(or(x,y)) -> not#(not(x))
p2: not#(and(x,y)) -> not#(y)
p3: not#(and(x,y)) -> not#(not(y))
p4: not#(and(x,y)) -> not#(x)
p5: not#(and(x,y)) -> not#(not(x))
p6: not#(and(x,y)) -> not#(not(not(x)))
p7: not#(or(x,y)) -> not#(y)
p8: not#(or(x,y)) -> not#(not(not(y)))
p9: not#(or(x,y)) -> not#(x)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: not#(or(x,y)) -> not#(not(x))
p2: not#(or(x,y)) -> not#(x)
p3: not#(or(x,y)) -> not#(not(not(y)))
p4: not#(or(x,y)) -> not#(y)
p5: not#(and(x,y)) -> not#(not(not(x)))
p6: not#(and(x,y)) -> not#(not(x))
p7: not#(and(x,y)) -> not#(x)
p8: not#(and(x,y)) -> not#(not(y))
p9: not#(and(x,y)) -> not#(y)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = x1 + (1,6)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,5)
          not_A(x1) = ((1,0),(1,1)) x1 + (0,2)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,5)
  
    precedence:
    
      or = and > not# = not
    
    partial status:
  
      pi(not#) = [1]
      pi(or) = []
      pi(not) = [1]
      pi(and) = []

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

p1: not#(or(x,y)) -> not#(not(x))
p2: not#(or(x,y)) -> not#(x)
p3: not#(or(x,y)) -> not#(not(not(y)))
p4: not#(or(x,y)) -> not#(y)
p5: not#(and(x,y)) -> not#(not(not(x)))
p6: not#(and(x,y)) -> not#(not(x))
p7: not#(and(x,y)) -> not#(not(y))
p8: not#(and(x,y)) -> not#(y)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: not#(or(x,y)) -> not#(not(x))
p2: not#(and(x,y)) -> not#(y)
p3: not#(and(x,y)) -> not#(not(y))
p4: not#(and(x,y)) -> not#(not(x))
p5: not#(and(x,y)) -> not#(not(not(x)))
p6: not#(or(x,y)) -> not#(y)
p7: not#(or(x,y)) -> not#(not(not(y)))
p8: not#(or(x,y)) -> not#(x)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = ((1,0),(1,1)) x1 + (1,9)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,8)
          not_A(x1) = x1 + (0,2)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,7)
  
    precedence:
    
      or = and > not > not#
    
    partial status:
  
      pi(not#) = [1]
      pi(or) = []
      pi(not) = [1]
      pi(and) = []

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

p1: not#(or(x,y)) -> not#(not(x))
p2: not#(and(x,y)) -> not#(y)
p3: not#(and(x,y)) -> not#(not(y))
p4: not#(and(x,y)) -> not#(not(x))
p5: not#(and(x,y)) -> not#(not(not(x)))
p6: not#(or(x,y)) -> not#(y)
p7: not#(or(x,y)) -> not#(x)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

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

p1: not#(or(x,y)) -> not#(not(x))
p2: not#(or(x,y)) -> not#(x)
p3: not#(or(x,y)) -> not#(y)
p4: not#(and(x,y)) -> not#(not(not(x)))
p5: not#(and(x,y)) -> not#(not(x))
p6: not#(and(x,y)) -> not#(not(y))
p7: not#(and(x,y)) -> not#(y)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = x1 + (0,10)
          or_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (1,13)
          not_A(x1) = ((1,0),(1,1)) x1 + (0,4)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,11)
  
    precedence:
    
      or = not = and > not#
    
    partial status:
  
      pi(not#) = [1]
      pi(or) = []
      pi(not) = [1]
      pi(and) = []

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

p1: not#(or(x,y)) -> not#(not(x))
p2: not#(or(x,y)) -> not#(x)
p3: not#(or(x,y)) -> not#(y)
p4: not#(and(x,y)) -> not#(not(not(x)))
p5: not#(and(x,y)) -> not#(not(x))
p6: not#(and(x,y)) -> not#(not(y))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: not#(or(x,y)) -> not#(not(x))
p2: not#(and(x,y)) -> not#(not(y))
p3: not#(and(x,y)) -> not#(not(x))
p4: not#(and(x,y)) -> not#(not(not(x)))
p5: not#(or(x,y)) -> not#(y)
p6: not#(or(x,y)) -> not#(x)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = ((0,0),(1,0)) x1 + (5,2)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (4,1)
          not_A(x1) = ((1,0),(1,1)) x1 + (0,2)
          and_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (4,5)
  
    precedence:
    
      not = and > not# > or
    
    partial status:
  
      pi(not#) = []
      pi(or) = []
      pi(not) = []
      pi(and) = []

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: not#(and(x,y)) -> not#(not(y))
p2: not#(and(x,y)) -> not#(not(x))
p3: not#(and(x,y)) -> not#(not(not(x)))
p4: not#(or(x,y)) -> not#(y)
p5: not#(or(x,y)) -> not#(x)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

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: not#(and(x,y)) -> not#(not(y))
p2: not#(or(x,y)) -> not#(x)
p3: not#(or(x,y)) -> not#(y)
p4: not#(and(x,y)) -> not#(not(not(x)))
p5: not#(and(x,y)) -> not#(not(x))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = ((1,0),(1,0)) x1 + (1,2)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (2,1)
          not_A(x1) = ((1,0),(1,1)) x1 + (0,6)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (2,3)
  
    precedence:
    
      not# = and > not = or
    
    partial status:
  
      pi(not#) = []
      pi(and) = []
      pi(not) = [1]
      pi(or) = []

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: not#(or(x,y)) -> not#(x)
p2: not#(or(x,y)) -> not#(y)
p3: not#(and(x,y)) -> not#(not(not(x)))
p4: not#(and(x,y)) -> not#(not(x))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

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: not#(or(x,y)) -> not#(x)
p2: not#(and(x,y)) -> not#(not(x))
p3: not#(and(x,y)) -> not#(not(not(x)))
p4: not#(or(x,y)) -> not#(y)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = x1 + (1,3)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,2)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,1)
          not_A(x1) = ((1,0),(1,1)) x1
  
    precedence:
    
      or = not > not# = and
    
    partial status:
  
      pi(not#) = [1]
      pi(or) = []
      pi(and) = []
      pi(not) = []

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: not#(or(x,y)) -> not#(x)
p2: not#(and(x,y)) -> not#(not(x))
p3: not#(and(x,y)) -> not#(not(not(x)))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

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: not#(or(x,y)) -> not#(x)
p2: not#(and(x,y)) -> not#(not(not(x)))
p3: not#(and(x,y)) -> not#(not(x))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = x1 + (1,6)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + (2,10)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + (2,8)
          not_A(x1) = x1 + (0,3)
  
    precedence:
    
      and > not > not# = or
    
    partial status:
  
      pi(not#) = [1]
      pi(or) = []
      pi(and) = []
      pi(not) = [1]

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: not#(or(x,y)) -> not#(x)
p2: not#(and(x,y)) -> not#(not(not(x)))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: not#(or(x,y)) -> not#(x)
p2: not#(and(x,y)) -> not#(not(not(x)))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = ((1,0),(1,0)) x1 + (2,9)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + (1,8)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + (1,7)
          not_A(x1) = x1 + (0,2)
  
    precedence:
    
      not > not# = or = and
    
    partial status:
  
      pi(not#) = []
      pi(or) = []
      pi(and) = []
      pi(not) = []

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: not#(and(x,y)) -> not#(not(not(x)))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: not#(and(x,y)) -> not#(not(not(x)))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          not#_A(x1) = ((0,0),(1,0)) x1 + (1,5)
          and_A(x1,x2) = ((1,0),(0,0)) x1 + (2,4)
          not_A(x1) = x1 + (0,1)
          or_A(x1,x2) = ((1,0),(0,0)) x1 + (2,4)
  
    precedence:
    
      not# = and = not > or
    
    partial status:
  
      pi(not#) = []
      pi(and) = []
      pi(not) = []
      pi(or) = []

The next rules are strictly ordered:

  p1

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