Rahul asked:

Suppose ‘no scientist are philosopher’ is true then which of these is true and which are false:

(i) no nonphilosopher are scientists
(ii) some nonphilosopher are nonscientist
(iii) all nonscientist are nonphilosopher
(iv) no scientist are nonphilosopher

Answer by Craig Skinner

The important thing here is to have a general method for solving this kind of predicate logic puzzle. Some people do it with the words, others with symbols, but it beats me how they manage.

Best is a diagram. Get pen and paper and follow these instructions (it’s much easier to do than the instructions suggest at first sight, I’d draw the diagram in this answer but my computer skills aren’t up to it).

* draw a square (this represents everything – philosophers, scientists, nonphilosophers, nonscientists)

* inside the square, draw a circle and put ‘P’ inside it – this circle is the philosophers

* also inside the square, draw another circle, not touching or overlapping the P circle, and put ‘S’ inside it – this circle is the scientists

* you can see that ‘no scientist is a philosopher’ is true (the circles don’t overlap)

* now, the part of the square outside P is obviously nonphilosophers, so put ‘nonP’ in the square somewhere away from the P circle

* note, S is also part of the square outside P, so write ‘nonP’ inside the S circle as well

* similarly, the part of the square outside S is nonscientists, so put ‘nonS’ in the square (beside the ‘nonP’ you wrote earlier will do)

* finally, put ‘nonS’ inside the P circle as well

So now you have a square labelled ‘nonP’/ ‘nonS’ inside which is a circle labelled ‘P’/nonS’ and a separate circle labelled ‘S/nonP’.

Now read off the answers for (i) – (iv).

(i) ‘no nonP are S’. FALSE. You can see that SOME nonP are S (the S circle)

(ii) ‘some nonP are nonS’. TRUE. the part of the square outside both circles is nonP/nonS

(iii) ‘all nonS are nonP’. FALSE. SOME nonS are P (the P circle)

(iv) ‘no S is nonP’. FALSE. ALL S are nonP (the S circle)