p: In the fisrt room there is a lady. |
$\neg$p: In the fisrt room there is a tiger. |
q: In the second room there is a lady. |
$\neg$q: In the second room there is a tiger. |
Translate given English statements into logical statement:
1. (p$\wedge\neg$q) 2. (p$\oplus$q)
As, given in the question, one of the sign is true and other is false. So, there are two cases possible.
Case 1: Suppose, First door's sign is True and Second door's is False.
1. (p$\wedge\neg$q) is True only if p is True and $\neg$q is True.
2. (p$\oplus$q) is False if both p & q are True Or False. but it's contradict with 1$^{st}$ statement. So, Case 1 is not valid.
Case 2: Suppose, First door's sign is False and Second door's is True.
1. (p$\wedge\neg$q) is False if atleast one of p & $\neg$q is False.
2. (p$\oplus$q) is True if either p is True & q is False or p is False & q is True. So, there are two possible subcases.
Subcase 1 : p is True and q is False.
but it's contradict with 1$^{st}$ statement. So, it's not valid.
Subcase 2: p is False and q is True.
it's not contradict with 1$^{st}$ statement. So, it's a valid subcase.
Hence, from subcase 2 of Case 2, we can conclude that Behind the Second door is the lady.