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Show that if a sentence φ of L1 contains no sentence letters, then φ is either a tautology or a contradiction. Use Lemma c, which states that:

If A(P) = B(P), for every sentence letter P in φ, then |φ|A = |φ|B

My attempt:

Towards a proof by contraposition, we can suppose that φ is not a tautology and φ is not a contradiction. This means that φ does not receive the same evaluation under all structures, or rather |φ|A ≠ |φ|B.

This is where I'm stuck: I'm tempted to say that if |φ|A ≠ |φ|B, φ must contain sentence letters (whereafter I could say that φ must contain sentence letters in order to receive different evaluations by different structures--this will have proved the contraposition). But this doesn't seem true -- constants can also have different evaluations under different structures.. is this the wrong strategy?

3

If the formula φ contains no sentence letters, then φ is either ⊤ or ⊥ or a truth-functional combination of them with connectives, like e.g. ⊤ → ⊥.

But ⊤ and ⊥ are sentential constants, in the sense that they have the same truth-value in every truth-assignment, i.e, for every A and B we have that A(⊤)=B(⊤) and A(⊥)=B(⊥).

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