# Logic for Computer Scientists/Propositional Logic/Equivalence and Normal Forms

## Equivalence and Normal Forms[edit | edit source]

Until now, we only discussed single formulae and their semantical
properties. In this section we start investigating whether formulae
can be transformed into another form, without changing their
semantics. For this we introduce the concept of *logical*
equivalence*, which can be used to investigate the transformation of*
a given formula into its *normal form*.

## Definition 6[edit | edit source]

The formulae and are called (semantically) equivalent, iff for all assignments for and , . We write .

Formulae containing different subformulae can be equivalent, e.g. all tautologies are equivalent. More interesting is the following theorem:

## Theorem 3 (Substituitivity)[edit | edit source]

Let and a formula which contains at least one occurrence of the subformula . Then it holds , where is obtained from by substituting any occurrence of by .

**Proof:** The proof is by induction over the structure of the
subformula :

Assume for the induction start, that is atomic, hence holds, and the result of substituting the only occurrence of by results in and because , we have . Assume the theorem holds for all proper subformulae of : If we have the same argumentation as above in the start of the induction. If we have three cases:

- : Because is a subformula of we can conclude that , where is constructed by substituting any occurrence of by . From the definition of the semantics of we conclude that and hence . Where H_1 is equivalent formula constructed by replacing F in H by G, resulting in H_1
- : Assume without loss of generality, that occurs in . Then, again we can conclude from the induction assumption, that and from the definition of the semantics of we conclude, that .
- is similar.

## Theorem 4[edit | edit source]

The following equivalences hold:

(Idempotence)

(Commutativity)

(Associativity)

(Absorption)

(Distributivity)

(Double negation)

(deMorgan's rule)

, if is a tautology

, if is a tautology (Rule of Tautology)

, if is unsatisfiable

, if is unsatisfiable (Rule of Unsatisfiability)

**Proof:** All equivalences can be proved by truth tables. This is
done here for the second rule of absorption:

() | ()) | ||
---|---|---|---|

false | false | ||

false | true | ||

true | false | ||

true | true |

Let us now use these equivalences together with the theorem of substituitivity (TS) to prove the following equivalence:

Associativity and TS

Commutativity and TS

Distributivity

Commutativity and TS

Distributivity and TS

Unsatisfiability and TS

Commutativity and TS

Commutativity and TS

## Problems[edit | edit source]

#### Problem 9 (Propositional)[edit | edit source]

The binary junctor for exclusive disjunction is defined by . Show that the following holds for propositional logic formulae , and :

- (Commutativity).
- (Associativity)

#### Problem 10 (Propositional)[edit | edit source]

Show the following by equivalence transformation. Give to all remodelling steps the used rules!

#### Problem 11 (Propositional)[edit | edit source]

Prove with equivalences:

- is a tautology.
- is unsatisfiable.

#### Problem 12 (Propositional)[edit | edit source]

The binary junctor with the meaning *neither-nor* is defined by . Let be
a propositional logic formula, which contains only the
operators and . Prove that for every
formula a formula exists, such that
and is built by using only the junctor .

#### Problem 13 (Propositional)[edit | edit source]

Let be a propositional logic formula, which contains only the operators
, and . Prove that for every
formula a formula exists, such that
and is built by using only the junctors and .

#### Problem 14 (Propositional)[edit | edit source]

The following formulae are given:

- Simplify by using .
- Simplify by using equivalences but not .

## Definition 7 (Normal Forms)[edit | edit source]

A literal is an atomic formula or the negation of an atomic formula (positive or negative, resp.) A formula is in conjunctive normalform (CNF) iff

where is a literal.

A formula is in disjunctive normalform (DNF) iff

where is a literal

## Theorem 5[edit | edit source]

For every formula there is an equivalent formula which is in DNF and an equivalent formula which is in CNF.

Let us formulate an algorithm to transform a given formula F into an equivalent normalform:

**Given:** **A formula **

1. Substitute in every subformula of the form

until there is no subformula of this kind.

2. Substitute in the result from the above step every subformula of the form

until there is no subformula of this kind.

**Result: An equivalent formula in CNF**

Until now, we investigate the transformation of a propositional formula into an equivalent normal form. Another problem in the context of normal forms is, to construct a normal form formula from a given truth table; i.e. the formula itself is not known, but its behaviour is given by a truth table. Let's read a normalform formula from a truth table: Assume a formula , which is given by the following truthtable.

A | B | C | F |
---|---|---|---|

false | false | false | true |

false | false | true | false |

false | true | false | false |

false | true | true | false |

true | false | false | true |

true | false | true | true |

true | true | false | false |

true | true | true | false |

In order to construct a formula in DNF, which is equivalent to , we have to take into account, that every line of the table which yields the truthvalue gives one conjunction: if the assignment of the literal is it is included as , if is we include . For the above example we get as a DNF:

If we change in the above procedure the roles of and
and and we arrive at a CNF:

We introduce a special representation for formulae in
normalform. In our circuit-example (circuit) from the introduction
we already used a very special form of normalforms, namely the
implication form for formulae in CNF: every subformula of a
conjunction is written as:

It is easy to see, that this implication is logically equivalent to a disjunction . Sometimes the implication is written as

Even the following ambiguous notation is used in some cases, where the comma in the premiss stands for a conjunction and the comma in the conclusion for a disjunction:

For some important procedures for logical reasoning it is mandatory to represent the formulae not only in one of the above notations for a normalform, moreover, it is necessary to use the so-called clause-form from the following definition.

## Definition 8[edit | edit source]

If is a formula in CNF, i.e.

then its corresponding representation in clause form is given as

The sets are called clauses.

This representation as sets of literals has the advantage that literals occur in no special order and that multiple occurrences of a literal in a disjunction are "merged" in its clause form. Note that as a consequence we have built in associativity, commutativity and Idempotence into the representation.

## Problems[edit | edit source]

#### Problem 15 (Propositional)[edit | edit source]

Create formulae in (a) conjunctive or (b) disjuncive normal form which are equivalent to:

where denotes exclusive or.

#### Problem 16 (Propositional)[edit | edit source]

The theorem prover OTTER uses the following optimization rules for clause sets:

**Subsumption:** If the literals of a clause are a subset of an another clause remove from the set of clauses.

**Deleting by unit clause:** If the set of clauses contains a unit clause -here is single literal unit-clause- every occurrence of a complementary literal in a clause is deleted.

Which laws of the logic justify these procedures?

#### Problem 17 (Propositional)[edit | edit source]

Generate truth tables for the following formulae. Give the conjunctive and disjunctive normal forms of the formulae. Which one of the relations , , and holds?

- whereas applies.

#### Problem 18 (Propositional)[edit | edit source]

Generate a CNF and a DNF form the following truth table for the formula .

A | B | C | F |
---|---|---|---|

0 | 0 | 0 | 0 |

1 | 0 | 0 | 1 |

0 | 1 | 0 | 0 |

1 | 0 | 0 | 1 |

1 | 1 | 0 | 1 |

1 | 0 | 1 | 0 |

0 | 1 | 1 | 1 |

1 | 1 | 1 | 0 |

#### Problem 19 (Propositional)[edit | edit source]

Generate a CNF from the formula