Looking for some help on Solving Syllogisms, just can't seem to grasp it.

Heres two problems

a. All smart asses are dumb asses.
b. Some bad asses are smart asses.

conclusions
1 All dumb asses are bad asses
2 Some dumb asses are bad asses
3 No bad asses are dumb asses
4 Some bad asses are no dumb asses

a. All ambulances are life savers.
b. No ambulances are bumper cars.


1 No life savers are bumper cars
2 No bumper cars are life savers
3 Some life savers are no bumper cars
4 None of the above

Easy to think of it in terms of Set Theory.

a. All smart asses are dumb asses
b. Some bad asses are smart asses

S = {smart asses}
D = {dumb asses}
B = {bad asses}

All S are D, so D encompasses S (S is a subset of D). Some B are in S. Not necessarily all B are in S, so not all can be said to be in D either.

Conclusion: 2. Since some B are in S, some B have to be in D since all of S is in D.

(Conclusion 4 cannot be determined since we don’t know the relation of B\S and D.)

—————————————————————————-

a. All ambulances are life savers
b. No ambulances are bumper cars

A = {ambulances}
L = {life savers}
B = {bumper cars}

All A are L, so L encompasses A (A is a subset of L). No A are in B. This does not give a relation for B and L though so it is unclear if any B can be found in L or vice versa.

Conclusion: 3. Some L are not in B. Since we know all of A are in L and none of A are in B, it follows that those elements of L which represent A are not in B.

This post was edited by timmayX on Feb 13 2018 08:31pm

thanks, and just another question. All the questions follow normal logic as well right? so for example, all humans have scales, which is obv invalid

I don’t think reality-based logic can be applied to these. Otherwise, “All smart asses are dumb asses” wouldn’t really hold water.

I think you need to observe the statements in a vacuum unless your instructor has advised that reality-based logic should be the first thing you consider.