YES

We show the termination of the TRS R:

  and(x,false()) -> false()
  and(x,not(false())) -> x
  not(not(x)) -> x
  implies(false(),y) -> not(false())
  implies(x,false()) -> not(x)
  implies(not(x),not(y)) -> implies(y,and(x,y))

-- SCC decomposition.

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

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

and R consists of:

r1: and(x,false()) -> false()
r2: and(x,not(false())) -> x
r3: not(not(x)) -> x
r4: implies(false(),y) -> not(false())
r5: implies(x,false()) -> not(x)
r6: implies(not(x),not(y)) -> implies(y,and(x,y))

The estimated dependency graph contains the following SCCs:

  {p3}


-- Reduction pair.

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

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

and R consists of:

r1: and(x,false()) -> false()
r2: and(x,not(false())) -> x
r3: not(not(x)) -> x
r4: implies(false(),y) -> not(false())
r5: implies(x,false()) -> not(x)
r6: implies(not(x),not(y)) -> implies(y,and(x,y))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          implies#_A(x1,x2) = max{x1 + 1, x2 - 3}
          not_A(x1) = max{6, x1 + 5}
          and_A(x1,x2) = max{8, x1 + 5}
          false_A = 4
    
      precedence:
      
        not > and = false > implies#
      
      partial status:
    
        pi(implies#) = []
        pi(not) = [1]
        pi(and) = [1]
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          implies#_A(x1,x2) = 0
          not_A(x1) = x1 + 2
          and_A(x1,x2) = x1 + 3
          false_A = 2
    
      precedence:
      
        and = false > implies# = not
      
      partial status:
    
        pi(implies#) = []
        pi(not) = [1]
        pi(and) = []
        pi(false) = []
    

The next rules are strictly ordered:

  p1

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