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Theorem 4.1 { A B } |- B A

Proof

A B A B Assumption

B A by E

B A by I

Theorem 4.2 |- A B B A

Proof

A B A B Assumption ( discharged by 4)

B A by E

B A by I

A B B A by I (4)

To deduce A B, it is often useful to make A as an assumption, with the

idea of discharging it later using the I rule.

Theorem 4.3 {B} |- A B

Proof

A Assumption ( discharged by 4)

B Assumption

A B by I (4)

However, there are no definite ways or guidelines to prove a theorem.

Analogously, there are no definite ways to write a computer program to

accomplish a specific task.

There could be many ways to prove the same theorem.

Thus, proving a theorem requires an experience and ingenuity.

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Theorem 4.4 |- B ( A B )

Proof

A Assumption (1)

B Assumption (2)

A B by I (from 1)

B (A B) by I (from 2)

Theorem 4.5 { A} |- ( A B )

Proof

A A Assumptions

by E

B by

A B by I, discharging A

Now we shall show that four derived rules can be obtained from the basic

rules.

Theorem 4.6 RAA is a derived rule

Proof

Suppose that we have a proof of from A.

A Assumption

by E

A by I, discharging A

A by E

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Theorem 4.7 Modus Tollens is a derived rule

Proof

A B Assumption

B Assumption

A Assumption

B by E

by E

A by I, discharging A

Theorem 4.8 -introduction is a derived rule

Proof

A Assumption

A Assumption

by E

A RAA, discharging A

Theorem 4.9 LEM is a derived rule

Proof

(P P) Assumption

P Assumption

P P by I

by E

P by I, discharging P

P P by I

by E

P P RAA, discharging (P P)

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Theorem 4.10 { A B, B } |- A

The presence of A B as an assumption suggests that we should try to deduce A

from A and from B and then to apply E to discharge both A and B.

Proof

B B Assumptions

A Assumptions and by E

A B A A Assumptions, Id and

A E, discharging A, B and A B

Theorem 4.11 |- ( A B) (A B)

Similarly, the presence of ( A B) as an assumption suggests that it might be

worth trying to deduce A B from both A and B.

Proof

A B Assumptions

A B A B A B Assump.and theorems 4.5 and 4.4

A B E, discharging A and B

( A B) (A B) by I, using A B

Theorem 4.12 ( A B) |- A

Instead of trying to prove ( A B) |- A directly, we will use A to get

A B by I, then introduce ( A B) and use the contradiction to get A. Proof

A Assumption

A B ( A B) I, Assumption

by E

A RAA, discharging A

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Theorem 4.13 (A B) |- ( A B)

Proof

A B Assumption

A A LEM

A Assumption

A B by I

A Assumption

B by E

A B by I

A B by E, discharging A and A