This is being done in the context of a Hilbert system in propositional logic.  But let's assume we've already proven all the axioms are tautologies / will always be true.  How then are we showing that modus ponens is also valid? (or is valid the wrong word to use here? Normally, "valid" for a statement means the logic holds IF the premises were true, but premises being true is what makes it actually sound... but when it comes to inference rules I don't know the terminology)  $P, (P \to Q) \vdash Q$  Is this strictly an informal "look at the truth tables and reason it out" kind of thing?  i.e. if we assume $P$ is true and $P \to Q$ is true because they are each reducible to either true axioms or true atomic propositions, then we know $P \to Q = \top$ becomes $\top \to Q = \top$. And in the truth tables for $\to$, this only has one solution, i.e. where $Q = \top$, therefore modus ponens holds? Or are we now showing that $P, (P \to Q) \vDash Q$?  I mean is that it? Or is it a more formal approach? I get totally lost when it comes to understanding and manipulating / proving things with these metalogical symbols.
123456
I have the following question: "Explain the concept of the soundness of a rule of a proof system with respect to the truth tables"  Would it be correct to state the following:  "A rule of a proof system is sound with respect to the truth tables iff the conditions of the rules (requirements of the rule) ensure that the final formula will behave according to it's intended truth table."?  This is a question in a practice exam, and I can not find a specific mentioning so far, by scanning Google and Proof Wiki.  What would the correct answer be?  Thank you.
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In the next proof we avail ourselves of the next lemma: For all sets A and B, ( A ∪ B ) - B = A. Proof: Let A and B be arbitrary sets and let x ∈ ( A ∪ B ) - B. x ∈ ( A ∪ B ) - B ⇔ x ∈ ( A ∪ B ) ∧ x ∉ B ⇔ ( x ∈ A ∨ x ∈ B ) ∧ x ∉ B ⇔ ( x ∈ A ∧ x ∉ B ) ∨ ( x ∈ B ∨ x ∉ B) For p ∧ ~ p ≡ F and p ∧ q ⇒ p where p and q are prepositions, ( x ∈ A ∧ x ∉ B ) ∨ ( x ∈ B ∨ x ∉ B ) ⇔ x ∈ A ∨ F By modus tollendo ponens, x ∈ A ∨ F ⇔ x ∈ A So then, x ∈ ( A ∪ B ) - B ⇔ x ∈ A And ( A ∪ B ) - B = A Therefore, the statement for all sets A and B, A ∩ B = ∅  implies ( A ∪ B ) - B = A is true trivially. Is this proof right?
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Provided we have this truth table where "$p\implies q$" means "if $p$ then $q$":  $$\begin{array}{|c|c|c|} \hline p&amp;q&amp;p\implies q\\ \hline T&amp;T&amp;T\\ T&amp;F&amp;F\\ F&amp;T&amp;T\\ F&amp;F&amp;T\\\hline \end{array}$$  My understanding is that "$p\implies q$" means "when there is $p$, there is q". The second row in the truth table where $p$ is true and $q$ is false would then contradict "$p\implies q$" because there is no $q$ when $p$ is present.  Why then, does the third row of the truth table not contradict "$p\implies q$"? If $q$ is true when $p$ is false, then $p$ is not a condition of $q$.  I have not taken any logic class so please explain it in layman's terms.       Administrative note. You may experience being directed here even though your question was actually about line 4 of the truth table instead. In that case, see the companion question  And even if your original worry was about line 3, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.
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I don't quite yet understand how $\oplus$ (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both. But when it comes to solving problems with them or proving equalities I have no idea how to use $\oplus$.  For example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not $A \oplus B = A \oplus C$  implies $B = C$ is true.   I know that the venn diagram of $\oplus$ in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this:    I feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the $\oplus$, but I'm not sure if that's an adequate enough proof.  On the other hand, I could be completely wrong about my reasoning.    Also just for clarity's sake: Would $A\cup B = A \cup C$ and $A \cap B = A \cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well.
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We have the following argument: $P \implies Q$ (Premise 1) $P$ (Premise 2) $—$ $∴Q$ (Conclusion)  The accompanying truth table is:    I don't understand this section of my textbook:     The premises are both true only in line four of the table, and in this   line the conclusion is true as well.      You can also see from the truth table that both premises are needed to   make this argument valid. But if we were to change the truth table for   the conditional statement to make P → Q false in the first line of the   table, then the second premise of this argument would no longer be   needed.       We would end up with the conclusion that, just from the single   premise P → Q, we could infer that Q must be true, since in the two   lines of the truth table in which the premise P → Q would still be   true, lines two and four, the conclusion Q is true too.   My thinking:  If we would change the $P → Q$ to be false in the first line of the truth table. Then why couldn't the conclusion be false as well? I don't understand what line $2$ and $4$ have anything to do with line $1$?
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This semester I take Introduction to logic and sets theory and for logic our reference is A Mathematical Introduction to Logic by H.B.Enderton.  I have some problems with this course I mean it is somehow unfamiliar ! our professor teach well and I learn at class but when I want to solve my homework ,I have no idea ,how to solve problems !  I don't have any idea how to prove this :  suppose a and b are tautologically equivalent, where A is wff and a is used in A and B is  another wff that b is used in; prove that A and B are tautologically equivalent.  Any one can help me ?
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I've been studying Discrete Mathematics from Kenneth Rosen's book. In that book, there is this "$(p \vee q) \land (\neg p \vee r)) \rightarrow (q \vee r)$" proposition if we can say so.   The book wants me to prove that this proposition is a tautology.  I've searched on the Internet and also on here but couldn't figure out how to prove this without a truth table. There is another asked question but that is so confusing and is not really illustrative.  Would you help me to prove this, please? Thank you in advance. :)
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When you have a propositional sentence of the form P &sup; Q&nbsp; &mdash; which we might read as "if P, then Q" &mdash; how can you tell when it is true, or false, based on the truth-values of P and Q in classical logic? When is this different from Q &sup; P? And what is the connection to 'necessary' conditions, or sufficient conditions?  Also: does P &sup; Q mean that P causes Q?
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I am trying to understand the fundamentals of mathematical logic in order to be able to study discrete mathematics and computer science soon.  I have a big problem understanding Implication. I understand the idea intuitively very well; for example:  $$ x &gt;0  \implies 2x&gt;0 .$$  Or  $$ x \text{ is a prime number} \implies x \geq 2 .$$  I notice that there is always a (connection) between the hypothesis and the conclusion; they seem to be related and fall in the same context.  What I don't understand is: according to the truth table, any two propositions can be linked through an implication even if they are not related at all, or they belong to different contexts.  For example :  $$ \text{A day is 24 hours} \implies \text{A cat has four legs and a tail} .$$  which is logically or mathematically TRUE, because both statements are true and according to the truth table when when both inputs are true for any two statements then the implication is true.  How can that be true?  Another example:  $$ 2 \text{ is a prime number} \implies \text{An hour is 60 minutes}.$$  again, which is logically or mathematically TRUE, because both statements are true and according to the truth table when when both inputs are true for any two statements, then the implication is true.  How can that be true? That's my first question.  The second question could be the same:     How can we use truth tables with implication anyway?   What I understand is truth tables are used to list the probabilities of the output based on the logical values of the inputs. So the value of the output of any line in the truth table depends (only) on the logical values of the inputs in that line and has nothing to do with the conditional connection between the two inputs.   How can we display a conditional statement in a truth table the same way we display a logic gate or so? In other words, how can I tell only from the values of $P$ and $Q$ that $P \implies Q$?
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we recently got the statement that $(P \land Q) \to P$ is a tautology, but i do not get why it is one, since it is wrong for certain cases of P and Q.
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I am reading a Set Theory book by Kunen. He presents first-order logic and claims that if a set of sentences in inconsistent, then it proves every possible sentence. Since he does not explicitly specify the inference rules, I became curious as to how fundamental this property of inconsistent systems is.  So my question is what is the simplest proof, with the least use of assumptions, of the vague claim that "inconsistent systems can prove anything" - in particular I'm interested in the assumptions about the system needed to prove this - is it true only for first order logic? Only for first order logic with the "standard" rules of inference (Modus ponens and GEN)? Or is it such a basic truth that it can be proved for every "reasonable" proof system (and what is "reasonable")?
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I would like to know if someone can explain in a somehow down to earth (almost logic free) way why is it true that in an axiom system where there is some statement $P$ such that $P$ and its negation $\lnot P$ are true, then every statement in the system is true?  I'm not sure if this can be done, but basically since I don't know any formal logic at all, I'm interested in seeing if at least the argument can be conveyed in an intuitive way, or if the idea can be explained without talking about first or second order logic and using symbols like $\top$, $\bot$, and $\vdash$.  This  is like the formal version (which I don't understand) so maybe my question can be thought of as a version for dummies of that question.   Thanks a lot in advance.
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I would just like some clarification on the use of the terms "theorem" and "tautology".  I have read  that a theorem is a "formula for which a zero-premise derivation has been provided".  But  it is a "logical consequence of the axioms".  These definitions seem to be in conflict, as are not axioms just premises (Formulae which are assumed to be true)?  And what about a tautology.  My understanding is that a tautology is the same as the first definition of a theorem.  Or is tautology a semantic thing (i.e. valid) whereas theorem is syntactic (i.e. derivable)?
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How do I prove $$S\rightarrow \exists xP(x) \vdash \exists x(S\rightarrow P(x))$$ using natural deduction? Just an alignment of which axioms or rules that one could use would be much appreciated.
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I am studying entailment in classical first-order logic.  The Truth Table we have been presented with for the statement $(p \Rightarrow q)\;$ (a.k.a. '$p$ implies $q$') is: $$\begin{array}{|c|c|c|} \hline p&amp;q&amp;p\Rightarrow q\\ \hline T&amp;T&amp;T\\ T&amp;F&amp;F\\ F&amp;T&amp;T\\ F&amp;F&amp;T\\\hline \end{array}$$  I 'get' lines 1, 2, and 3, but I do not understand line 4.      Why is the statement $(p \Rightarrow q)$ True if both p and q are False?  We have also been told that $(p \Rightarrow q)$ is logically equivalent to $(~p || q)$ (that is $\lnot p \lor q$).  Stemming from my lack of understanding of line 4 of the Truth Table, I do not understand why this equivalence is accurate.       Administrative note. You may experience being directed here even though your question was actually about line 3 of the truth table instead. In that case, see the companion question  And even if your original worry was about line 4, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.
84831
Can you answer me , please.  1- Show that (p→r)∧(q→r) and (p∨q)→r are logically equivalent  2- show that (p → r) ∨ (q → r) and (p ∧ q) → r are logically equivalent  without truth table .
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I'm currently studying logic and I'm confused about what is meant by the truth or falsity of statements within an argument. From Daniel J. Velleman's, How to Prove It, pg.9:  &quot;Although we have no guarantee the conclusion is true, it can only be false if at least one of the premises is false ... We will say an argument is valid if the premises cannot all be true without the conclusion being true as well.&quot;  Having then read a  about the validity of an argument, I am told the following:  &quot;The following argument is of the same logical form but with false premises and a false conclusion, and it is equally valid: All cups are green.  Socrates is a cup. Therefore, Socrates is green.&quot;   I feel that there is some distinction being made here between the terms &quot;truth&quot; and &quot;false&quot; in the two instances that I'm not quite grasping. In the wikipedia article I believe &quot;false&quot; refers to the general truth of each individual statement i.e. we know all cups are not green in reality so this is a &quot;false&quot; statement. However in the Velleman book I am not sure what he means by the &quot;truth&quot; or &quot;falsity&quot; of the premise and conclusion. In sentential logic what is meant by the &quot;truth&quot; or &quot;falsity&quot; of a statement – the &quot;empirical truth&quot; of the statement, or does it refer to some other notion of &quot;truth&quot; and &quot;falsity&quot;?