Additional Implication Laws and Derivations as Trees

The following two laws handle conjunctions that occur either as premises or as conclusions.

The first, the conjunction-premise law, handles the case where the conjunction is one of the premises; the second, conjunction-conclusion law, handles the case where the conjunction is the conclusion. We adopt the following notation:

(∧, |=): for the conjunction-premise law, (|=, ∧): for the conjunction-conclusion law.

Here are the two laws:

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

(∧, |=) follows, via (10) (and monotonicity), from the obvious facts that (i) A ∧ B logically implies each of A and B, and (ii) A, B |= A ∧ B.

In the second law, (|=, ∧), the ⇒-direction follows, via transitivity, from the fact that A ∧ B implies each of A and B. The ⇐-direction (which is the direction we will be using in top-down derivations) is established by noting that, if Γ|= A and Γ |= B, then Γ implies every premise in the list A, B. Since this list implies A∧ B, Γ implies it as well (via transitivity).

The conjunction-conclusion law is distinguished from the other laws used so far in that it reduces a goal (Γ|= A ∧ B) not to one but to two goals. Both must be achieved. The number of goals increases thereby. It may increase further, since each of the two new goals may give rise, directly or indirectly, to more than one goal. But the new goals are simpler: instead of the conclusion A∧B we have only A or B. This is also true of the other laws that we shall use.

It constitutes the main feature of the method: Although the number of goals may increase, the goals themselves become simpler. In the end, the initial goal is reduced to a collection of so called elementary goals; these are goals whose validity can be immediately checked.

Here is a simple top-down derivation that ends with two final, self-evident implications:

1. C → A |= (C → B) → (C → A∧B)

2. C → A, C → B, C |= A ∧ B two applications of (|=, →), 3. C, A, B |= A ∧ B two applications of disjoining,

4.1 C, A, B |= A by (|=, ∧), √

4.2 C, A, B |= B by (|=, ∧). √

Note: In (∨, |=), like in (|=, ∧), we get (via the ⇐ direction) a reduction of a goal to two goals, both of which must be proved. In (|=, ∧) the English ‘and’ on the right-hand side corresponds to ∧ in the conclusion. But in (∨, |=) it corresponds to ∨ in the premise. Some students find this confusing. The reason for converting ∨ to ‘and’ is that ∨ occurs in the premise. In order to show that ‘... or ’ implies something, we have to show that ‘...’

implies it and ‘ ’ implies it.

(|=, ∨) can be proved by replacing A ∨ B by the equivalent ¬A → B and, via (|=, →), transferring ¬A to the premises.

Homework 4.4 Give an argument that proves the disjunction-premise law.

Note: To show that the premises imply A∨B, it is sufficient to show that they either imply A or imply B (because each of A, B, implies A∨ B). Hence, we can have a disjunction-conclusion law with ‘or’ in the right-hand side. But such a law holds only in the ⇐-direction.

The ⇒-direction does not hold in general. Γ may imply A ∨ B without implying any of the disjuncts A, B. For example,

|= A ∨ ¬A

(here Γ is empty), but from this it does not follow that either |= A, or |= ¬A. For if A is neither logically true nor logically false, then 6|= A and 6|= ¬A.

Therefore you run some risk if, in order to show that Γ|= A ∨ B, you try to show that either Γ |= A or Γ |= B. For if Γ implies neither disjunct, you are sure to fail, even though Γ may imply the disjunction. For example, you will fail if you try to prove in this way that

|= A ∨ ¬A. On the other hand, trying to show that Γ |= A ∨ B, by showing that Γ, ¬A |= B, is safe; for the second task is equivalent to the first.

Here is an example of a derivation that involves more than one branching.

1. A∨ B |= (A → B) → [(B → A) → A∧B],

2. A∨ B, A → B, B → A |= A∧B two applications of (|=, →),

3.1 A, A→ B, B → A |= A∧B by (∨, |=),

3.2 B, A→ B, B → A |= A∧B by (∨, |=),

4.1 A, B, B→ A |= A∧B by disjoining,

5.11 A, B, B→ A |= A by (|=, ∧), √

5.12 A, B, B→ A |= B by (|=, ∧), √

4.2 A→ B, B, A |= A∧B by disjoining,

5.21 A→ B, B, A |= A (|=, ∧), √

Note: If the conclusion is (A∧B)∧C, then two applications of (|=, ∧) will reduce our initial goal to three, each with one of A, B, C as a conclusion. We may consider a law that achieves it in one step. Here it is convenient to disregard the grouping in the repeated conjunction:

Γ|= A ∧ B ∧ C ⇐⇒ Γ |= A and Γ |= B and Γ |= C

The same applies to longer conjunctions. Repeated disjunctions in the premises can be treated similarly. Such laws introduce branching into more than two branches. They are not included among our basic laws.

The conditional-premise and conditional-conclusion laws, denoted respectively as (→, |=) and (|=, →), are as follows:

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

The first is obtained by replacing the premise A→ B by the equivalent ¬A ∨ B and applying the disjunction-premise law. The second is our old friend (7) (with the two sides of ⇔ reversed).

Note: If, in addition to A→ B, the premise-list contains A, we can apply disjoining:

Γ, A, A→ B |= C ⇐⇒ Γ, A, B |= C

This is better than applying (→, |=). But whereas (→, |=) is always applicable to a premise that is a conditional, disjoining requires that the antecedent, A, be among the premises.

It is good practice to apply disjoining as long as the premises contain a conditional and its antecedent; e.g., a list of premises of the form

Γ, A→ B, B → C, A

can be reduced, by two applications of disjoining, to the equivalent and much simpler list Γ, A, B, C

Sometimes (12^∗) can be applied as well; e.g., relying on the obvious implication A|= A ∨ B, we can replace Γ, A, (A∨ B) → C by the equivalent premise-list: Γ, A, C.

The remaining binary connective, ↔, can be dealt with through replacing A ↔ B by the conjunction of two conditionals: (A→ B) ∧ (B → A). Alternatively, we can employ directly the following laws:

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

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

(↔, |=) is obtained by replacing A ↔ B by the equivalent (A∧B) ∨ (¬A∧¬B) and then applying the premise laws for disjunction and conjunction. (|=, ↔) is obtained by replacing the biconditional by a conjunction of two conditionals and applying the conclusion laws for conjunction and conditional.

Treatment of Negated Compounds Negated sentential compounds are sentences either of the form ¬(¬A), or of the form ¬(A B), where is a binary connective. If a negated compound occurs either in the premises or as the conclusion, then, if it is ¬(¬A) the goal can be simplified by dropping the double negation. In the other case, we can push negation inside, using our old equivalence laws. This yields a sentence to which we can apply one of the previous implication laws.

If is ∧ or ∨, the pushing inside is done via De Morgan’s laws. If it is →, the pushing inside is achieved via the equivalence:

(13) ¬(A → B) ≡ A ∧ ¬B

In the case of biconditional we can use each of the equivalences:

(14.i) ¬(A ↔ B) ≡ (A∧¬B) ∨ (¬A∧B)

(14.ii) ¬(A ↔ B) ≡ A ↔ ¬B

(If the negated biconditional is a premise, then, usually, (14.i) is more convenient; we then apply (∨, |=) and, to each of the resulting implications, we apply (∧, |=). If the negated bicon-ditional is the conclusion, then (14.ii) is more convenient; we then apply (|=, ↔). Sometimes (14.ii) is more convenient when¬(A ↔ B) is a premise: if the premise-list contains A, we can replace A↔ ¬B, A by the equivalent list A, ¬B.)

Homework

4.5 Derive (14.ii) from (14.i) by pushing-in disjunction, deleting redundant conjuncts and replacing the remaining disjunctions by equivalent conditionals.

4.6 Establish the following implications via top-down derivations. You can use all the laws introduced so far, as well as substitution of equivalents. At the end you should have reduced the initial goal to a bunch of self-evident goals in which the conclusion is among the premises.

(In some cases you might have to use (12^∗)).

1. |= ¬(A → B) → A 2. |= A∧B → (A ↔ B)

3. A∨ B, A → C |= ¬B → C 4. (A→ B) → A |= A

5. |= [C → (¬A → B)] → [(C → A) ∨ (C → B)]

6. (A∨ B) → A∧B |= A ↔ B

7. [A→ (A ↔ B)] ∧ [B → (A ↔ B)] |= A ↔ B 8. |= A → (B → C) ↔ (A∧B → C)