I'm sure someone has thought of this before, but I haven't seen this justification (if it is one at all) for why the truth-table of the material conditional is the way it is in both literature on the subject, and on the internet.

The justification goes as follows - two points seem to make intuitive sense:

(a) The biconditional 'A↔B' is true if and only if both A and B have the same truth-value.

(b) The biconditional 'A↔B' is the same thing as saying '(A→B)&(B→A)'

Together with the truth-table for the conjunction, we get the unique truth-table the material conditional has to have for these two criteria to hold.

Does this heuristic justification hold any water?



You're saying that for the biconditional to have the truth table that it does, and for it to be equivalent to (A_B)&(B_A), then _ has to be → with its current truth table. But there are other options:

A   B   A_B   (A_B)&(B_A)
T   T    T         T
T   F    T         F
F   T    F         F
F   F    T         T

A   B   A_B   (A_B)&(B_A)
T   T    T         T
T   F    F         F
F   T    F         F
F   F    T         T
  • Thanks for pointing out this error. Do you know of any way to fix my argument in one way or another? Jul 23 '16 at 8:01
  • 3
    There is a way to derive material implication from inferential considerations. If you posit a two place truth function P * Q, then constrain it so that modus ponens is valid, i.e. P * Q; P preserves truth to Q, then constrain it so that modus tollens is valid, i.e. P * Q; ¬Q preserves truth to ¬P, then constrain it so that affirming the consequent is invalid, i.e. P * Q; Q does not preserve truth to P, you can show that the only truth table for P * Q that is consistent with these constraints is material implication.
    – Bumble
    Jul 23 '16 at 10:31
  • @Bumble that was extremely insightful and interesting, thank you a lot, but I am still interested in an ad hoc assumption, call it (c), that would make my argument go through. In other words, I need to add an assumption which would decide between the actual truth-table of the material implication and the two truth-tables EliranH has described. Jul 23 '16 at 11:13
  • 1
    The two counterexamples in Eliran's response are B→A and A↔B respectively. You can eliminate the latter by requiring (c) A→B does not entail B→A. To eliminate the former you need something that breaks the symmetry between A→B and B→A. You could use (d) A→B together with A entails B.
    – Bumble
    Jul 24 '16 at 8:59

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