The Basic Laws

The method uses a finite number of basic laws. Some were mentioned before and some are easily obtained from previously mentioned laws. Not all laws considered above are taken as basic. The basic laws are naturally classified as follows.

First, there is a law that enables trivial rearrangements of premise lists:

If every sentence occurring in Γ occurs in Γ⁰ and every sentence occurring in Γ⁰ occurs in Γ, then for every A,

Γ |= A ⇐⇒ Γ⁰ |= A

Using this law we can reorder the premises, delete repetitions, or list any premise more than once. Henceforth, such reorganizing will be carried out without explicit mention.

Next, we designate two types of implications as self-evident:

Self-Evident Implications

Γ, A |= A Γ, A, ¬A |= B

Implications belonging to these two types play a role similar to that of axioms. In bottom-up proofs they serve as starting points. In top-down ones they are the final successful goals.

The rest, referred to as reduction laws, are the basis of the method. They enable us to replace a goal by simpler goals. The first group consists of laws that handle sentential compounds A B. For each binary connective, , we have a premise law ( ,|=) and a conclusion law (|= ), which handle, respectively, -compounds that appear as a premise, or as the conclusion. The second group, which deals with negated compounds, is presented later.

Laws for Conjunction

(∧, |=) Γ, A∧ B |= C ⇐⇒ Γ, A, B |= C

(|=, ∧) Γ |= A ∧ B ⇐⇒ Γ |= A and Γ |= B

Laws for Disjunction

(∨, |=) Γ, A∨ B |= C ⇐⇒ Γ, A |= C and Γ, B |= C

(|=, ∨) Γ |= A ∨ B ⇐⇒ Γ,¬A |= B

Laws for Conditional

(→, |=) Γ, A→ B |= C ⇐⇒ Γ,¬A |= C and Γ, B |= C

(|=, →) Γ |= A → B ⇐⇒ Γ, A |= B

Laws for Biconditional

(↔, |=) Γ, A↔ B |= C ⇐⇒ Γ, A, B |= C and Γ, ¬A, ¬B |= C (|=, ↔) Γ |= A ↔ B ⇐⇒ Γ, A |= B and Γ, B |= A

The goal-reduction process is as described in 4.3. To recap: a step consists of replacing the left-hand side of a reduction law (the goal that is being reduced) by the right-hand side.

The ⇐-direction guarantees that proving the new goals (or goal) is sufficient for proving the old goal; the ⇒-direction means that they are also implied by it. All counterexamples to a new goal are also counterexamples to the old one, and any counterexample to the old one is obtained in this way.

A goal’s children are the goals that replace it. (If there is one goal on the right, there is only one child). It follows from the above that if the children are valid so is the parent.

Consequently, if all the leaf goals are valid, so are their parents, and the parents of their parents, and so on, up to the initial goal. On the other hand, any counterexample to one of the leaf goals is also a counterexample to one (or more) of their parents, hence also to the parent’s parent, and so on, up to the initial goal. And any counterexample to the original goal is a counterexample to a goal in some leaf.

We still need laws for reducing negated compounds, sentences of the form ¬¬A or ¬(A B).

The laws for double negation allow us to drop it, either in a premises or in the conclusion.

Compounds of the form ¬(A B) can be treated in the way described in 4.2.2 (page 124), i.e., by pushing the negation inside. This means that we use laws such as:

Γ, ¬(A∧B) |= C ⇐⇒ Γ, ¬A∨¬B |= C

And similar laws for pushing negation inside in ¬(A∨B), in ¬(A → B), and in ¬(A ↔ B).

There is a more elegant way: Combine in a single law the pushing of negation and the law that applies to the resulting compound. For ¬(A ∧ B) this yields:

Γ, ¬(A∧B) |= C ⇐⇒ Γ, ¬A |= C and Γ, ¬B |= C .

Doing so for all connectives, we get reduction laws for negated compounds, of the same type as our previous ones. It is easy to see that counterexample equivalence is also true for the second group. Because these laws are obtained from the first group by substituting sentential expressions by logically equivalent ones; such substitutions do not change the sets of counterexamples.

Laws for Negated Negations

(¬¬, |=) Γ, ¬¬A |= B ⇐⇒ Γ, A |= B

(|=, ¬¬) Γ |= ¬¬B ⇐⇒ Γ |= B

Laws for Negated Conjunctions

(¬∧, |=) Γ, ¬(A ∧ B) |= C ⇐⇒ Γ, ¬A |= C and Γ, ¬B |= C

(|=, ¬∧) Γ |= ¬(A ∧ B) ⇐⇒ Γ, A |= ¬B

Laws for Negated Disjunctions

(¬∨, |=) Γ, ¬(A ∨ B) |= C ⇐⇒ Γ, ¬A, ¬B |= C

(|=, ¬∨) Γ |= ¬(A ∨ B) ⇐⇒ Γ, |= ¬A and Γ |= ¬B

Laws for Negated Conditionals

(¬ →, |=) Γ, ¬(A → B) |= C ⇐⇒ Γ, A, ¬B |= C

(|=, ¬ →) Γ |= ¬(A → B) ⇐⇒ Γ |= A and Γ |= ¬B

Laws for Negated Biconditionals

(¬ ↔, |=) Γ, ¬(A ↔ B) |= C ⇐⇒ Γ, A, ¬B |= C and Γ, ¬A, B |= C (|=, ¬ ↔) Γ |= ¬(A ↔ B) ⇐⇒ Γ, A |= ¬B and Γ, ¬B |= A

This completes the set of reduction laws.

Branching Laws: A law whose right-hand side has more than one implication is called a branching law. Each application of a branching law causes branching in the tree. The branching laws are, for non-negated compounds: (|=, ∧), (∨, |=), (→, |=), (↔, |=) and (|=, ↔) . For negated compounds they are: (¬∧, |=), (|=, ¬∨), (|=, ¬ →), (¬ ↔, |=) and (|=, ¬ ↔). The other laws are referred to as non-branching.

Memorization: You do not have to memorize all the laws. A useful strategy is to memorize only four: the two laws for conjunction, the premise-disjunction law, (∨, |=), and the conclusion-conditional law, (|=, →). The rest you can get by obvious substitutions of equivalents: The conclusion-disjunction law–by rewriting A∨ B as ¬A → B; the premise-conditional law–by rewriting A→ B as ¬A∨B; the premise-biconditional law–by rewriting the biconditional as (A∧ B) ∨ (¬A ∧ ¬B), and the conclusion-biconditional law–by rewriting it as (A → B) ∧ (B → A). Beside the double negation laws, the other laws for negated compounds are obtained by pushing negation in, as indicated earlier.

Elementary Implications

Elementary implications are those that cannot be simplified through reduction laws. An implication is elementary if every sentential expressions figuring in it is either a sentential variable or a negation of one. This is equivalent to saying that it contains neither binary nor negated compounds. Here are some examples.

A, ¬B, C, ¬D |= ¬B ¬A, C, B, ¬C |= D ¬A, B, ¬C, D |= ¬E Claim: (I) If an elementary implication is valid, then it is self-evident, i.e., either the conclusion occurs as a premise or the premises contain a sentential expression and its negation.

(II) If an elementary implication is not self-evident then there is a unique assignment to its sentential variables that constitutes a counterexample. This assignment is determined by the following conditions:

(i) Every sentential variable that occurs unnegated in the premises gets T, and every sentential variable that occurs negated in the premises gets F.

(ii) The sentential variable of the conclusion gets F, if it occurs unnegated, T–if it occurs negated.

Proof: Assume that an elementary implication is not self-evident and show that the condi-tions in (II) determine an assignment, and that the assignment is the unique counterexample.

For any given assignment the following is obvious: All the premises get T iff the assignment satisfies (i). The conclusion gets F iff the assignment satisfies (ii). There is at most one assignment, to the sentential variables occurring in the implication, that satisfies (i) and (ii);

because (i) and (ii) prescribe truth-values to all these sentential variables. Hence there is a counterexample iff there is an assignment satisfying (i) and (ii); the counterexample is then unique.

The only way in which (i) and (ii) can fail to determine an assignment is by prescribing more than one truth-value for the same sentential variable. This does not happen unless the

implication is self-evident. For if it is not, no sentential variable occurs in the premises both negated and unnegated; hence (i) assigns to each sentential variable occurring in the premises exactly one value. Next, if the variable of the conclusion does not occur in the premises, then only (ii) gives it a value. If it occurs in the premises, it must be either negated in the premises and unnegated in the conclusion, or unnegated in the premises and negated in the conclusion.

Otherwise the conclusion is among the premises and the implication is self-evident. Hence (ii) and (i) assign to it the same value.

QED

In order to check the validity of an elementary implication, we therefore check if it is self-evident. If it is not, then (i) and (ii) in (II) tell us what the counterexample is.

Of the three elementary implications given above, the first two are self-evident. The coun-terexample to the third is:

A B C D E

F T F T T

We can now assemble all the pieces and sum up the method.