GATE CSE 2002 | Question: 1.8

Question

"If $X$ then $Y$ unless $Z$" is represented by which of the following formulas in propositional logic? ("$\neg$" is negation, "$\land$" is conjunction, and "$\rightarrow$" is implication)

• A. $(X\land \neg Z) \rightarrow Y$
• B. $(X \land Y) \rightarrow \neg Z$
• C. $X \rightarrow(Y\land \neg Z)$
• D. $(X \rightarrow Y)\land \neg Z$

Comments

good trick sir ...thanks lakshman sir

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it helped me thanks..!

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$A \text{  }\underline{\text{unless}}\text{  } B \equiv A + B$

$\therefore ($If $X$ then $Y)$ unless $Z \equiv ($If $X$ then $Y) + Z \equiv X’  + Y + Z$

$\equiv$ If $X$ then $(Y$ unless $Z) \equiv $If $X$ then $(Y + Z) \equiv X’  + Y + Z$

 

Option A: $(X \land \neg Z) \rightarrow Y = (XZ')' + Y = X' + Z + Y =$ (Ans.)

Option B: $(X \land Y) \rightarrow \neg Z = X' + Y' + Z'$

Option C: $X \rightarrow (Y \land \neg Z) = X' + YZ'$

Option D: $(X \rightarrow Y) \land \neg Z = (X' + Y)Z' = X'Z' + YZ'$

Answer 1

The statement “If X then Y unless Z” means, if Z doesn’t occur, X implies Y i.e. ¬Z→(X→Y), which is equivalent to Z∨(X→Y) (since P→Q ≡ ¬P∨Q), which is then equivalent to Z∨(¬X∨Y). Now we can look into options which one matches with this.

So option (a) is (X∧¬Z)→Y = ¬((X∧¬Z))∨Y = (¬X∨Z)∨Y, which matches our expression. So option A is correct.

Answer 2

Option A is correct.

2nd interpretation....it is saying that if X happens then Y will happen otherwise Z will happen..So can we say that it means X and Z can not happen simultaneously because if X happens then output will be be Y not Z so can we also say that X happen and Z does not happen then output will be Y

( X ^ ~Z ) -> Y

Answer 3

The statement “If X then Y unless Z” means, if Z doesn’t occur, X implies Y i.e. ¬Z→(X→Y), which is equivalent to Z∨(X→Y) (since P→Q ≡ ¬P∨Q), which is then equivalent to Z∨(¬X∨Y). Now we can look into options which one matches with this.

So option (a) is (X∧¬Z)→Y = ¬((X∧¬Z))∨Y = (¬X∨Z)∨Y, which matches our expression. So option A is correct.

Answer 4

      X →(¬Z→Y)

⟹ ¬X ∨ (Z ∨ Y)

⟹ (¬ X ∨ Z) ∨ Y

⟹ ¬(X ∧ ¬Z) ∨ Y

⟹ (X ∧ ¬Z) → Y  So , (A)