Re: Derivations

My professor has reduced the number of SD rules to eight, deleting the
rules of biconditional and the rule of reiteration because they are
redundant.

BTW, there is no need to use vertical lines in proofs. And the rules
should be written in metalogical symbols.

Using double negation as the rule of negation elimination is better.
Anyway, I will use your rule of negation elimination for demonstration.
I will demonstrate some of your problems. C and F are left to you to
exercise your skills.

a) {A v B, ~B}├ A

1) A v B PREMISS
2) ~B PREMISS
3) A (ASSUMPTION) Discharged
4) B (ASSMP) Discharged
5) ~A (ASSMP) Discharged
6) B & ~B 2,4 CONJ. INTRO.
7) A 5-6 NEGATION ELIMINATION
8) A (3,4)-7 DISJUCTION ELIMINATION


b) {A→(~B→C), A & ~B}├ C v E

1) A→(~B→C) Premiss
2) A & ~B Premiss
3) A 2, Conj. Elimination
4) ~B 2, Conj. Elimination
5) ~B→C 1, 3, Conditional Elimination
6) C 5, 4, Conditional Elimination
7) C V E 6, Disjuction Intro.


c) {(~A v ~B)→C, D & ~C}├ A

d) {A → ~~B, C → ~B}├ ~(A & C)

1) A → ~~B Premiss
2) C → ~B Premiss
3) A & C (Assumption) Discharged.
4) A 3, Conj. Elimination
5) C 3, Conj. Elimination
6) ~B 2, 5 Condictional Elimination
7) ~~B 1, 4 Condictional Elimination
8) ~B & ~~B 6, 7 Conj.
9) ~(A & C) 3-8 Negation Intro.

e) A →(B → A)

1) A (Assumption) Discharged
2) B (Assumption) Discharged
3) B→A 2,1 Conditional Intro. (There is no need to
introduce the rule of reiteration)
4) A→(B→A) 1,3 Conditional Intro.

f) ~A → ((B & A) → C)

g) (A v B) → (B v A)

1) A V B (Assumption) Discharged
2) A (Assumption) Discharged
3) B V A 2, Disjuction. Intro
4) B (Assumption) Discharged
5) B V A 4, Disj. Intro
6) B V A 2-5 Disjuction Elimination
7) (A v B) → (B v A) 1-6 Conditional Intro.

h) A ←→ ~~A

A)
1) A (Assumption) Discharged
2) ~A (Assumption) Discharged
3) A & ~A 1, 2 Conj. Intro
4) ~~A 2-3 Negation Intro.
5) A → ~~A 1-4 Conditional Intro.

B) 1) ~~A (Assumption) Discharged
2) ~A (Assumption) Discharged
3) ~A & ~~A 1,2 Conj. Intro
4) A 2-3 Negation Intro.
5) ~~A → A 1-4 Conditional Intro.

Combine Section A and B by the rule of bicondictional.

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