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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      implies#_A(x1,x2) = ((1,0,0),(1,0,0),(1,1,0)) x1 + ((1,0,0),(1,1,0),(1,0,1)) x2
      not_A(x1) = x1 + (3,1,1)
      and_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((0,0,0),(0,0,0),(1,0,0)) x2 + (4,4,5)
      false_A() = (2,5,8)

The next rules are strictly ordered:

  p1

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