Lecture Notes, 13 Sep 2010

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1. Show that (p → (q → r)) ≡ ((p ∧ q) → r):

   To show that 2 propositions are logically equivalent,
   we need to show that their truth tables have the same input-output (I-O) columns:

   T-table for (p → (q → r)):

   p	q	r	(q → r)	(p → (q → r))
   T	T	T	T	T
   T	T	F	F	F
   T	F	T	T	T
   T	F	F	T	T
   F	T	T	T	T
   F	T	F	F	T
   F	F	T	T	T
   F	F	F	T	T

   T-table for ((p ∧ q) → r):

   p	q	r	(p ∧ q)	((p ∧ q) → r)
   T	T	T	T	T
   T	T	F	T	F
   T	F	T	F	T
   T	F	F	F	T
   F	T	T	F	T
   F	T	F	F	T
   F	F	T	F	T
   F	F	F	F	T

   Note that the I-O columns are the same, even though the intermediate columns are different.

   ---

2. One topic of CSE 191 is learning how to do proofs. Here's one:

   Theorem:

   Let A, B be propositions.
   Then A ≡ B iff (A ↔ B) is a tautology.

   I.e., two propositions (A, B) are equivalent iff their biconditional (A ↔ B) is a tautology.

   Remarks:

   1. Note that the book gives this as the definition of "logically equivalent".
      This kind of thing happens frequently in math:
      One mathematician's definition is another's theorem
      (and, whenever that's the case, the first mathematician's theorem is the second mathematician's definition)!

   2. Note that there are 3 different uses of a kind of "biconditional" in this theorem:
      • The "≡" says that two propositions are logically equivalent.
      • The "iff" says that two mathematical-English sentences
        ("A ≡ B" and "(A ↔ B) is a tautology") are necessary and sufficient for each other.
      • The "↔" connects A with B to form a molecular proposition.

   proof:

   We have to show that a sentence of the form "P iff Q" is the case,
   where P = "A ≡ B"
   and where Q = "(A ↔ B) is a tautology".

   To do that, we need to show two things:

   1. If P, then Q
   2. If Q, then P
   1. To show: if P, then Q
      we need to assume (or suppose, or make believe) that P is the case
      and then show that—under that assumption (or under that supposition, or in that hypothetical situation)—Q is the case

      Suppose A ≡ B, & show (A ↔ B) is a tautology:

      Because A ≡ B, we can replace "≡" with its definition:
      For all rows of the t-table, tval(A) = tval(B)

      But, if we now add a column for (A ↔ B) to the t-table,
      it will get filled in with Ts, because of the semantics of ↔

      So, for all rows, tval(A ↔ B) = T.

      But this is the definition of a tautology!

      So, (A ↔ B) is a tautology
      which is what we wanted to prove ("QED") in this first part of the proof

      [This is where we ended today, but, for the sake of completeness, here's the rest of the proof, which we'll review in class on Wednesday.]

   2. To show: if Q, then P
      we need to assume that Q, and then show that P:

      Suppose (A ↔ B) is a tautology, & show A ≡ B:

      Because (A ↔ B) is a tautology, we can replace "is a tautology" by its definition:
      for all rows, tval(A ↔ B) = T

      But the only way for tval(A ↔ B) = T is for tval(A) = tval(B), by the semantics of ↔

      So, for all rows, tval(A) = tval(B)

      But this is the definition of logical equivalence!

      So, A ≡ B,

      which is QED for the second part of the proof.

   Therefore, QED for the complete proof.