Question

Find the proof of the following

((a ∧ b) ∨ (c ∧ d)), (a → e), (b → f), (c → f), (d → e) ⊢ e

Answer 1

¬ (not)

∧ (and)

∨ (or)

→ (implies)

↔ (if and only if)

The

meaning can be calculated using the standard truth tables.

A B ¬A A ∧ B A ∨ B A → B A ↔ B

1 1 0 1 1 1 1

1 0 0 0 1 0 0

0 1 1 0 1 1 0

0 0 1 0 0 1 1

Double negation ¬(¬p) ≡ p
Excluded middle p ∨ ¬p ≡ Tr ue
Contradiction p ∧ ¬p ≡ F al se
Identity laws Tr ue ∧ p ≡ p
F al se ∨ p ≡ p
Idempotent laws p ∧ p ≡ p
p ∨ p ≡ p
Commutative laws p ∧ q ≡ q ∧ p
p ∨ q ≡ q ∨ p
Associative laws (p ∧ q)∧r ≡ p ∧(q ∧r )
(p ∨ q)∨r ≡ p ∨(q ∨r )
Distributive laws p ∧(q ∨r ) ≡ (p ∧ q)∨(p ∧r )
p ∨(q ∧r ) ≡ (p ∨ q)∧(p ∨r )
DeMorgan’s laws ¬(p ∧ q) ≡ (¬p)∨(∧q)
¬(p ∨ q) ≡ (¬p)∧(∧q)
Conjunction introduction. If you know A, and you know B, you can conclude A ∧B.
Conjunction elimination If you know A ∧B, then A.
Negation elimination If ¬(¬A) is true, then so is A.
Negation introduction If you assumed A, and you arrived at a contradiction,
then you can say ¬A is true.
Disjunction introduction If A is true, then A ∨B is true.
Disjunction elimination If you know A ∨B, and you know B is false, then A is true.
Implication introduction If you assume A, and arrive at B, you can conclude A → B.
Modus Ponens If you know A → B is true, and you know A is true,
you can conclude B.
Modus Tollens If you know A → B, and you know ¬B,
you can conclude ¬A.
Law of syllogism If you know A → B, and you know B →C,
you can conclude A →C.
Universal elimination If you know ∀xP(x), you can conclude P(c)
for an arbitrary c in the universe of discourse.
Universal introduction If you know P(c) for an arbitrary c
in the universe of discourse, you can conclude ∀xP(x).
Existential introduction If you know P(c) for a specific c
in the universe of discourse, you can conclude ∃xP(x).
Existential elimination If you know ∃xP(x), you can conclude P(c)
for a specific c in the universe of discourse.

By using this you can prove these results.