0

If A entails B, why is it not true that A is also consistent with B? Given the definition requires A and B to be able to be true at the same time.

2
  • 2
    If A is itself inconsistent then it entails any proposition, including B. In which case, A and B are not consistent.
    – Bumble
    Apr 4, 2019 at 20:30
  • 1
    "The definition requires A and B to be able to be true at the same time", definition of what?
    – Conifold
    Apr 4, 2019 at 20:37

3 Answers 3

2

If A is consistent and A entails B, then {A, B} is consistent.

If A is inconsistent, then A entails B, but {A, B} is not consistent.

2

We can clarify a little bit the relationship between some fundamental logical concepts using their definitions.

In classical logic we have that A entails B (or : B is a logical consequence of A) iff A → B is valid (in propositional logic : is a tautology).

But a formula is valid iff its negation is unsatisfiable.

Thus, A entails B iff ¬ (A → B) is unsatisfiable.

In turn, the formula ¬ (A → B) is unsatisfiable iff the set of formulas { A, ¬B } is inconsistent.

Conclusion :

A entails B iff { A, ¬B } is inconsistent.

1

It may help to rephrase things as, "A entails B iff B is 'consistent relative to A:'" if A entails B, then any consistent set of sentences containing A remains consistent when we add B to it.

Entailment - like lots of mathematical notions - allows vacuity: if A is inconsistent, then since there are no models of A at all, it is vacuously true that every model of A is also a model of B ("all the elephants in this room are pink"). So this is why the "relative to A" clause above is important. Alternatively, we could say "If A entails B, then either {A,B} is consistent or the entailment is 'stupid' (= A entails B by virtue of A being false)."

1
  • 1
    For ease of readability I'm conflating a bit between "consistency" and "satisfiability" in the above; they are in fact equivalent, but this is a deep fact and I feel obliged to mention the issue. Apr 5, 2019 at 18:57

You must log in to answer this question.

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