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I'm having trouble proving this tautological consequence. I'd hope that you guys can maybe oversee my process and identify errors, because I went over this couple of times and I arrive at the same conclusion.

The question goes like this:

$A \Rightarrow B$

$C \Rightarrow B$

Therefore: $(A \lor C) \Rightarrow B$

Q: Show that the conclusion of the arguments is a tautological consequence of the premises using truth tables.

This is how I tried to solve it:

My solution to the problem. It seems like I'm missing something very fundamental here.

Natalie Clarius
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Vocaloidas
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  • You seem to confusing deduction rules (as used in proofs) with tautologies (statements that are true in all truth assignments). You have decribed a proof rule not a tautology and your attempt tries to show the conclusion is a tautology, which it is not (by itself). – Henno Brandsma Dec 21 '16 at 05:34

2 Answers2

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The problem is really:

A implies B

C implies B is

(A implies B) and (C implies B)

And then it should be easily provable

edit:

\begin{array}{ccc|c@{}ccc@{}ccccc@{}ccc@{}ccc@{}ccc@{}c} a&b&c&(&a&\lor&c&)&\rightarrow&b&\leftrightarrow&(&a&\rightarrow&b&)&\land&(&c&\rightarrow&b&)\\\hline 1&1&1&&1&1&1&&1&1&\mathbf{1}&&1&1&1&&1&&1&1&1&\\ 1&1&0&&1&1&0&&1&1&\mathbf{1}&&1&1&1&&1&&0&1&1&\\ 1&0&1&&1&1&1&&0&0&\mathbf{1}&&1&0&0&&0&&1&0&0&\\ 1&0&0&&1&1&0&&0&0&\mathbf{1}&&1&0&0&&0&&0&1&0&\\ 0&1&1&&0&1&1&&1&1&\mathbf{1}&&0&1&1&&1&&1&1&1&\\ 0&1&0&&0&0&0&&1&1&\mathbf{1}&&0&1&1&&1&&0&1&1&\\ 0&0&1&&0&1&1&&0&0&\mathbf{1}&&0&1&0&&0&&1&0&0&\\ 0&0&0&&0&0&0&&1&0&\mathbf{1}&&0&1&0&&1&&0&1&0& \end{array}

SAJW
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  • Why did you rearrange C Implies B is (A implies B) and (C implies B). Am I missing a rule/law here? – Vocaloidas Dec 21 '16 at 05:06
  • The former are two seperated statements, but I assume you want to check wether if both statements are true, is then (AvC)>B true? Otherwise it makes simply no sense. @Vocaloidas – SAJW Dec 21 '16 at 05:08
  • This makes sense. Thank you very much. – Vocaloidas Dec 21 '16 at 05:10
  • Ah, just tried this solution. It doesn't really work. Fails on the third row where A => B is False and C =>B is False. Following the and rule, the result should also be False. – Vocaloidas Dec 21 '16 at 05:26
  • @Vocaloidas see edit. it is true, so make a new table – SAJW Dec 21 '16 at 05:47
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make a truth table of $$((a \rightarrow b) \land (c \rightarrow b)) \rightarrow ((a \lor c) \rightarrow b)$$

instead.

Henno Brandsma
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  • Just had a quick re-read on the chapter about tautologies and contradiction, now it's all as clear as a day. I arrived at the same proposition and when tested it all came out as a tautology. Thank you for your answer, i'll mark it as the answer. – Vocaloidas Dec 21 '16 at 06:46