5.2 Venn Diagrams


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Course lecture I developed over section 5.2of Patrick Hurley\\\'s "A Concise Introduction to Logic".

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5.2 Venn Diagrams

  1. 1. 5.2 Venn Diagrams
  2. 2. Venn Diagrams <ul><li>Venn diagrams are used again here to determine the validity of syllogisms, except they encompass the middle term, as well as the subject and predicate terms. </li></ul><ul><li>Seven quadrants are labeled in this diagram, each representing a different combination of S, M, and P. </li></ul>
  3. 3. Rules for testing syllogisms <ul><li>Marks on the diagrams (either shading or drawing an X) can only be done for the premises, not the conclusion. </li></ul><ul><li>If a universal claim is in the premises, it should be entered first into the diagram. If there are two universals, either one can be inserted first. </li></ul><ul><li>When inspecting a completed diagram to check for validity, make sure to remember that with particular claims, to say that “Some S are P” means that at least one S exists and that S is a P. “Some S are not P” means at least one S exists and that S does not fall within the circle that refers to P. </li></ul><ul><li>When shading in an area, be sure to shade the entire area and not just part. </li></ul>
  4. 4. Rules, continued <ul><li>Any area where you can draw is an X is initially divided into two parts. If one part is shaded, the X always goes in the unshaded part. </li></ul><ul><li>If one part is not shaded (meaning both are unshaded), then draw the X on the line that divides the two sections, since you know that the X falls within that area, but you don’t know which exact quadrant. (Whether one or the other, or even both). </li></ul>
  5. 5. Procedure for performing a validity test <ul><li>Start by inserting the claims in order from universals to particulars. </li></ul><ul><li>Then check to see if the claim made in the conclusion is supported by what is in the diagram. </li></ul><ul><li>If it is, the syllogism is valid. </li></ul><ul><li>If not, then it’s invalid. </li></ul>
  6. 6. Venn diagrams and the Boolean standpoint <ul><li>Remember, Boolean diagrams don’t have existential import (no X’s in the universal claims), so they’re easier to test. </li></ul><ul><ul><li>Example: </li></ul></ul><ul><ul><ul><li>No P are M. (EAE-2) </li></ul></ul></ul><ul><ul><ul><li>All S are M. </li></ul></ul></ul><ul><ul><ul><li>--------------------- </li></ul></ul></ul><ul><ul><ul><li>Therefore, no S are P. </li></ul></ul></ul><ul><li>No parts of the S circle are left unshaded that overlap with P, so this syllogism is unconditionally valid. (Because it’s valid from the Boolean standpoint) </li></ul>
  7. 7. Boolean, continued <ul><li>Example: </li></ul><ul><ul><li>All M are P. (AEE-1) No S are M. ------------------- </li></ul></ul><ul><ul><li>Therefore, no S are P. </li></ul></ul><ul><li>The conclusion claims that no part of S is in P (that the whole area where they overlap is empty) but here we see that one quadrant is not shaded in. So this syllogism is invalid. </li></ul>
  8. 8. Boolean, continued <ul><li>Example: </li></ul><ul><ul><li>Some M are P (IAI-1) </li></ul></ul><ul><ul><li>All S are M. </li></ul></ul><ul><ul><li>------------------- </li></ul></ul><ul><ul><li>Therefore, some S are P. </li></ul></ul><ul><li>The conclusion claims that at least one S exists inside the P circle (that there is an X in the overlap) but here we see the X is halfway in and halfway out. </li></ul><ul><li>It’s not clear whether the X is inside or outside the overlap, so it’s invalid. </li></ul>
  9. 9. Boolean, continued <ul><li>Example: </li></ul><ul><ul><li>Some M are not P. (OIO-1) </li></ul></ul><ul><ul><li>Some S are M. </li></ul></ul><ul><ul><li>---------------------- Therefore, some S are not P. </li></ul></ul><ul><li>This one is unique since there are no universal claims, thus no shading is involved. </li></ul><ul><li>Both claims in the premise involve marking down an X but we don’t know where exactly each one is in relation to the third term, so it’s invalid. </li></ul>
  10. 10. Aristotelian diagrams <ul><li>First, we find the form of a syllogism (like with the Boolean, by finding the mood and figure) then testing it from the Boolean. If it’s valid, then you’re done, since it’s valid from both standpoints. </li></ul><ul><ul><li>If it’s not valid, test it from the Aristotelian. </li></ul></ul><ul><li>To test from the Aristotelian, see if there’s a circle that is completely shaded except for one area, then insert an X in that area and retest it. </li></ul><ul><ul><li>If it contains no area that is shaded except for one area, then it’s invalid. </li></ul></ul><ul><li>If the form is conditionally valid, then see if the X represents something that actually exists. If it does, then it’s valid from the Aristotelian standpoint. </li></ul><ul><ul><li>If the X is something that doesn’t exist, then it’s invalid. </li></ul></ul>
  11. 11. Aristotelian, continued <ul><li>Example: </li></ul><ul><ul><li>No fighter pilots are tank commanders. (EAO-3) </li></ul></ul><ul><ul><li>All fighter pilots are courageous individuals. </li></ul></ul><ul><ul><li>---------------------------- </li></ul></ul><ul><ul><li>Therefore, some courageous individuals are not tank commanders. </li></ul></ul><ul><li>First, we reduce it to its form. </li></ul><ul><ul><li>No F are T. </li></ul></ul><ul><ul><li>All F are C. </li></ul></ul><ul><ul><li>---------------------------- </li></ul></ul><ul><ul><li>Therefore, some C are not T. </li></ul></ul>
  12. 12. Aristotelian, continued <ul><li>First, we test it from the Boolean. </li></ul><ul><li>The conclusion claims there is an X inside of C but outside of T, but there is not. So this syllogism is invalid from the Boolean standpoint. </li></ul><ul><li>Next we try testing it from the Aristotelian. </li></ul>
  13. 13. Aristotelian, continued <ul><li>There is a circle that is shaded except for one area (F) so we insert an X then retest it, to see if the conclusion is supported. </li></ul><ul><li>The conclusion asserts that there will be an X inside of C but outside of T. That requirement has been met, so the syllogism is conditionally valid. </li></ul><ul><li>The X stands for F (since we inserted it inside of the F circle) and we know fighter pilots exist. Therefore, this syllogism is valid. </li></ul>