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$(A → B) → C, A ∧ B \vdash C$

1.$\hspace{1cm}(A → B) → C \hspace{1cm}$premise

2.$\hspace{1cm}A ∧ B \hspace{2.5cm}$ premise

$\hspace{2cm}$ 3. $\hspace{1cm} A \to B \hspace{1cm}$ Assumption

$\hspace{2cm}$ 4. $\hspace{1cm} A \hspace{2cm} ∧ e_1 \hspace{0.5cm} 2$

$\hspace{2cm}$ 5. $\hspace{1cm} B \hspace{2cm} \to e \hspace{0.5cm} 3 $

6.$\hspace{1.75cm} C \hspace{2.5cm} \to e \hspace{0.5cm} 3, 1 $

I know this is wrong and I have tried multiple things but all of them feel wrong. My problem is that we already have A and B are true so I can't derive B by assuming A because of both are true. If I can't do that then how am I supposed to get C from the first premise. Also given A and B are true, I can't say that $A \to B$ , can I?

Mauro ALLEGRANZA
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Need_MathHelp
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  • From 2nd premise by $(\land \text E)$ you have $A$ and $B$ separately. From $B$, you have $(A \to B)$ by $(\to \text I)$. – Mauro ALLEGRANZA Oct 24 '23 at 11:30
  • But doesn't implication introduction rule state that $(A \to B)$ when we get or assume A and then somehow can arrive at B, then $(A \to B)$. At least that is how I understood it? In other words either I have not understood the rule or I don't understand why you are able to use $(\to I)$. – Need_MathHelp Oct 24 '23 at 11:49
  • See the post Can you Derive C→A from A alone, given the introduction rule? – Mauro ALLEGRANZA Oct 24 '23 at 11:57
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    This user has just posted 8 really feeble homework questions in the last two or three days. I suggest we stop answering until they give some sign of actually having worked through a text, and got a basic understanding. – Peter Smith Oct 24 '23 at 13:43
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    @PeterSmith I'm not doing homeworks, I am studying for an exam and I never actually ever asked for answers in my questions. I have always asked what is wrong with my solution. Some of the questions I am doing don't have answers so I am left to think whether my solution is correct or not so I come here to veirfy. Just on my previous question, I even asked a way I can verify my solutions on my own. – Need_MathHelp Oct 24 '23 at 14:00
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    @Need_MathHelp OK. If you look at https://logicmatters.net/ifl you will find you can freely download my Intro to Formal Logic which clearly explains both a set of Natural Deduction rules and (what you seem to need) proof-strategies for using them. AND there are extensive worked answers to the exercises. I suggest reading the ND chapters carefully before doing the exercises. And then you can check your answers with all the suggested worked answers. – Peter Smith Oct 24 '23 at 14:38
  • What are the exact rules of your system? There are pretty much as many different rule sets as there are textbooks in logic :P – Bram28 Oct 24 '23 at 16:45
  • @Bram the only rules I am allowed to use are Introductions and Eliminations of and, or, negation, implication and a couple of other derived formulas such as LEM, MT, PBC to name a few. – Need_MathHelp Oct 24 '23 at 18:25

1 Answers1

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You can derive the conditional $A \to B$ by a technique Paul Tomassi refers to as augmentation on p. $64$ of his book Logic. First, place the second conjunct $B$ of the formula $A \wedge B$ on a line by itself via conjunction elimination (see line $3$ below). Next, assume the antecedent $A$ of the conditional $A \to B$ you want to derive (line $4$ below). Then, use conjunction introduction and conjunction elimination to derive the formula $A \wedge B$ yet again and once more place $B$ on a line by itself (lines $5$ and $6$ below). In doing so, you construct a subproof in which $B$ (on line $6$) follows from the assumption of $A$ (on line $4$), so you can now discharge the assumption and derive the conditional $A \to B$ via arrow introduction (line $7$ below).

$ \begin{array}{llll} \{1\} & 1. & (A \to B) \to C & \text{premise} \\ \{2\} & 2. & A \wedge B & \text{premise} \\ \{2\} & 3. & B & \text{2 $\wedge e_2$} \\ \{4\} & 4. & A & \text{Assumption} \\ \{2,4\} & 5. & A \wedge B & \text{3,4 $\wedge i$} \\ \{2,4\} & 6. & B & \text{5 $\wedge e_2$} \\ \{2\} & 7. & A \to B & \text{4,6 $\to i$} \\ \{1,2\} & 8. & C & \text{1,7 $\to e$} & \square \\ \end{array} $

Once you have the conditional $A \to B$, the remainder of the proof in which you derive $C$ is rather straightforward (see line $8$).

RyRy the Fly Guy
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  • Yes but in page 64 and as well as the example Mauro gave earlier are not exactly the same. For example in 64, it was never said P was true so therefore you could assume P and get to the conclusion Q is true. But in my case A is true according to my second premise so isn't it a little weird and illogical to assume something that is already true? – Need_MathHelp Oct 24 '23 at 18:23
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    It is perfectly fine to assume something is true (regardless of whether or not it is actually true) if the purpose is to show that something else is a logical consequence of the assumption. Think of it this way... if it is raining outside, then is it illogical for me to say "if it is raining outside, then it will be wet outside?" No. The point of deriving a conditional such as $A \to B$ via arrow introduction is to show that some proposition $B$ follows whenever another proposition $A$ is true. Whether or not the antecedent $A$ is in fact true is irrelevant. – RyRy the Fly Guy Oct 24 '23 at 18:45
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    Yes now I get it when you say it like that. Thank you so much, I really appreciate all your help lately! – Need_MathHelp Oct 24 '23 at 18:46
  • One more question, but in your solutions, can we not just skip step 3 and 5. And then derive B from the premise and the assumption A? – Need_MathHelp Oct 25 '23 at 10:29
  • it's going to depend on the preference of your teacher. i have seen proofs where those steps are skipped because the author and audience are all aware of the implicit reasoning involved. Regardless, it is important to know that those particular steps are taking place in the background and help justify the conclusions being drawn. – RyRy the Fly Guy Oct 25 '23 at 15:39