# 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

Option A
Explanation
: In (I) we have to check conclusion between E and B. So see which propositions do we need as per the FIRST & LAST RULE. We have B in two statement, (a) and (b) while we have E in proposition (d). Note that the common entity between (b) and (d) is C. So we need just (b) and (d)
Solve in the order (d)+ (b) so that FIRST & LAST RULE is followed. No + ALL= Some Not reversed. Some B are not E. So (I) is true.
(II) we need to find relation between B and D. We need only proposition (b) and (c) for this
Take Reverse (c) + (b) = Some D are C + All C are B= Some D are B. Hence (II) is false

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

Option D
Explanation
: For (I): (a) + reverse(b) = All + Some  No Conclusion between A and C. So I is false
(II): reverse(b)+(c)= Some+ Some= No conclusion between B and D. So II is false

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

Option A
Explanation
: For (I): (a)+(b)= Some+All= Some = Some A are C is true. So I follows
(II): (a)+(b)+(c)+reverse(d)= Some+All+Some+Some  => solve one by one in mind.. Some+All= Some then add it to next Some, Some +Some No conclusion. Stop here. So there is no definite conclusion between A and E. So II doesnot follows.

Topic for Day 5: All the cases of possibility and Condition of Complementary Pair (EITHER-OR)