Shortcut to solve syllogism by rules
Day 4: Solving 3 or more statements in one go by rules
Example: Some A are B + All B are C + All C are D.
Method 1: Take first 2 proposition: We have Some+ All = Some
Then this Some with the last proposition i.e Some +All= Some.
Hence We have Some A are D is the conclusion.
Method 2: Take 2nd and 3rd proposition i.e All+ All = All
Now solve it with first proposition Some+All=Some => Some A are D
In both case we get the same answer. But there are some cases where you have to follow a sequence in solving the proposition to get the conclusion. Let’s have a look on the next example.
We will discuss again about this irregularity and a special case related to it on the last day.
Solved Examples:
1) Statements: (a) Some A are B. (b) All C are B. (c)Some C are D. (d)No E is C.
Conclusions:
I. Some B are not E.
II. No B is D
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow
2) Statements: All A are B. Some C are B. Some C are D. Some E are C.
Conclusions:
I. Some A are C
II. No B is D
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow
3) Statements: Some A are B. All B are C. Some B are D. Some E is D.
Conclusions:
I. Some A are C
II. No E is A
A) only I follows
B) only II follows
C) either I or II follows
D) neither I nor II follow
E) both I and II follow
Topic for Day 5: All the cases of possibility and Condition of Complementary Pair (EITHER-OR)
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Thank You!!!!!