YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

and R consists of:

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

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:
          not#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          and_A(x1,x2) = x1 + ((0,0),(0,1)) x2 + (1,1)
          or_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
  
    precedence:
    
      not# > and > or
    
    partial status:
  
      pi(not#) = []
      pi(and) = []
      pi(or) = [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: not#(and(x,y)) -> not#(x)
p2: not#(or(x,y)) -> not#(x)
p3: not#(and(x,y)) -> not#(y)

and R consists of:

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

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

and R consists of:

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

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:
          not#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          and_A(x1,x2) = x1 + ((0,0),(0,1)) x2 + (1,1)
          or_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
  
    precedence:
    
      and = or > not#
    
    partial status:
  
      pi(not#) = []
      pi(and) = []
      pi(or) = []

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

and R consists of:

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

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

and R consists of:

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

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:
          not#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          and_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
          or_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
  
    precedence:
    
      not# = and > or
    
    partial status:
  
      pi(not#) = []
      pi(and) = [2]
      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)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

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: and#(x,or(y,z)) -> and#(x,y)
p2: and#(x,or(y,z)) -> and#(x,z)

and R consists of:

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

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:
          and#_A(x1,x2) = ((1,0),(0,0)) x2
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,1)
  
    precedence:
    
      or > and#
    
    partial status:
  
      pi(and#) = []
      pi(or) = []

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: and#(x,or(y,z)) -> and#(x,y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: and#(x,or(y,z)) -> and#(x,y)

and R consists of:

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

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:
          and#_A(x1,x2) = ((1,0),(0,0)) x2
          or_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(1,1)) x2 + (1,1)
  
    precedence:
    
      and# = or
    
    partial status:
  
      pi(and#) = []
      pi(or) = []

The next rules are strictly ordered:

  p1

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