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?

1 Answer 1



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, 2016 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, 2016 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, 2016 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, 2016 at 8:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .