Solutions to Chapter 47 Proving equivalences

A. Consider the following sentences:

  1. 1.
    ​

    (A→¬B)

  2. 2.
    ​

    ¬(A↔B)

  3. 3.
    ​

    (¬A∨¬(A∧B))

  4. 4.
    ​

    (¬(A→B)∧(A→C))

  5. 5.
    ​

    (¬(A∨B)↔((¬C∧¬A)→¬B))

  6. 6.
    ​

    ((¬(A∧¬B)→C)∧¬(A∧D))

For each sentence, find an equivalent sentence in DNF and one inCNF by giving a chain of equivalences. Use (Id), (Absorp), and (Simp) to simplify your sentences as much as possible. We give a solution for (2). Removing ‘↔’ and pushing negations inward is common to both:

¬(A↔B)
¬((A→B)∧(B→A))
Bicond
¬((¬A∨B)∧(B→A))
Cond
¬((¬A∨B)∧(¬B∨A))
Cond
¬(¬A∨B)∨¬(¬B∨A)
DeM
(¬¬A∧¬B)∨(¬¬B∧¬A)
DeM
(A∧¬B)∨(¬¬B∧¬A)
DN
(A∧¬B)∨(B∧¬A)
DN
The result is now in DNF. To obtain a CNF, we keep going, using (Comm) and (Dist):
((A∧¬B)∨B)∧((A∧¬B)∨¬A)
Dist
(B∨(A∧¬B))∧((A∧¬B)∨¬A)
Comm
((B∨A)∧(B∨¬B))∧((A∧¬B)∨¬A)
Dist
((B∨A)∧(B∨¬B))∧(¬A∨(A∧¬B))
Comm
((B∨A)∧(B∨¬B))∧((¬A∨A)∧(¬A∨¬B))
Dist
The result can be simplified using (Simp):
(B∨A)∧((¬A∨A)∧(¬A∨¬B))
Simp
(B∨A)∧(¬A∨¬B)
Simp