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33 votes

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edited
Jan 21, 2017
by One

Tautology: *A tautology is a proposition that is always ture. *

*ex: (p v ~p)= T*

Contradiction: *A contradiction is a proposition that is always false.*

*ex: (p ^ ~p)=F*

Contigency: *A contigency is a proposition that is neither a tautology nor a contradiction.*

*ex: (p v q)----> ~r*

* A propositional logic is said to be satisfiable if it is either a tautology or contigency.

* If logic is contradiction then it is said to be unsatisfiable.

* By contigency we means the logic can be ture or false.

* Contradiction is the unsatisfiable function.

* Statement is valid means tautology.

* Statement is not valid means not tautology.

42

2 votes

Given statement $(P \Rightarrow (Q \vee R)) \Rightarrow ((P \wedge Q) \Rightarrow R)$ can be simplified as follows

$ \equiv (\neg P \vee (Q \vee R) ) \Rightarrow (\neg (P \wedge Q) \vee R) \\ \equiv (\neg P \vee R \vee Q ) \Rightarrow (\neg P \vee \neg Q \vee R) \\ \equiv ((\neg P \vee R )\vee Q ) \Rightarrow ((\neg P \vee R) \vee \neg Q) \\ $

Let $(\neg P \vee R ) = A $, then $(A \vee Q ) \Rightarrow (A \vee \neg Q) $ [simplifying this further]

$ \neg(A \vee Q ) \vee (A \vee \neg Q) $

$ (\neg A \wedge \neg Q) \vee (A \vee \neg Q) $

$[(\neg A \wedge \neg Q) \vee A] \vee \neg Q $ [As, $ X \vee (\neg X \wedge Y) = X \vee Y $ ]

$ (\neg Q \vee A) \vee \neg Q $

$ (\neg Q \vee A) $

$ (\neg Q \vee (\neg P \vee R ) ) $

For $Q=F$, result is True

For $Q=T, P=T, R=F$, result is False

Hence, Satisfiable(true atleast once) but not valid

$ \equiv (\neg P \vee (Q \vee R) ) \Rightarrow (\neg (P \wedge Q) \vee R) \\ \equiv (\neg P \vee R \vee Q ) \Rightarrow (\neg P \vee \neg Q \vee R) \\ \equiv ((\neg P \vee R )\vee Q ) \Rightarrow ((\neg P \vee R) \vee \neg Q) \\ $

Let $(\neg P \vee R ) = A $, then $(A \vee Q ) \Rightarrow (A \vee \neg Q) $ [simplifying this further]

$ \neg(A \vee Q ) \vee (A \vee \neg Q) $

$ (\neg A \wedge \neg Q) \vee (A \vee \neg Q) $

$[(\neg A \wedge \neg Q) \vee A] \vee \neg Q $ [As, $ X \vee (\neg X \wedge Y) = X \vee Y $ ]

$ (\neg Q \vee A) \vee \neg Q $

$ (\neg Q \vee A) $

$ (\neg Q \vee (\neg P \vee R ) ) $

For $Q=F$, result is True

For $Q=T, P=T, R=F$, result is False

Hence, Satisfiable(true atleast once) but not valid

0

1 vote

Check for validity by making given implication $false$

Option $\large A$

THE BELOW ANSWER IS WRONG

By Exportation Law,

$P \implies Q \implies R \equiv (P\land Q ) \implies R$

applying it to given proposition

$\left(P \implies \left(Q \vee R\right)\right) \implies \left(\left(P \wedge Q \right)\implies R\right) \equiv \left( P \land \left( Q \lor R\right) \land \left( P \land Q\right) \implies R \right)$

checking validity first

(P⟹(Q∨R))⟹((P∧Q)⟹R)

≡ (P∧(Q∨R)∧(P∧Q)⟹R)

As per my understanding, we can not write as per above using that exportation law.

@subbus

Because in question brackets are given as below

(P⟹(Q∨R)) ⟹ ((P∧Q)⟹R)

≡ ( (P⟹(Q∨R)) ∧ (P∧Q) ) ⟹ R.

(A → B) → (C → D) this is in question.

You have taken this as

A → (B → (C → D))

≡ (A ^ B) → (C → D)

≡ ((A ^ B) ^ C) → D

this differs from original question.

Please, check and correct me if wrong.

≡ (P∧(Q∨R)∧(P∧Q)⟹R)

As per my understanding, we can not write as per above using that exportation law.

@subbus

Because in question brackets are given as below

(P⟹(Q∨R)) ⟹ ((P∧Q)⟹R)

≡ ( (P⟹(Q∨R)) ∧ (P∧Q) ) ⟹ R.

(A → B) → (C → D) this is in question.

You have taken this as

A → (B → (C → D))

≡ (A ^ B) → (C → D)

≡ ((A ^ B) ^ C) → D

this differs from original question.

Please, check and correct me if wrong.

1

you are right @rajankakaniya.

My mistake is that I thought the implication like other logical connectives is associative but it turns out that implication is only right-associative.

Thank you for commenting :). I shall correct my answer now

1

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