I want to prove the law of excluded middle:  $A$  Thus $B ∨ ¬B$  under Copi's logic system.('A' means any premise. I want to derive the law of excluded middle from any premise, not means no premise) Copi system is constructed by 19 rules.  9 rules of inference: modus ponens(MP), modus tollens(MT), hypothetical syllogism(HS), disjunctive syllogism(DS), constructive dilemma(CD), simplification(Simp), conjunction(Conj), addition(Add), absorption(Abs)  10 rules of replacement: De Morgan's rule(DM), Commutativity(Com), Associativity(Assoc), Distrivution(Dist), Double Negation(DN), transposition(Trans), Material implication(Impl), Material equivalence(Equiv), exportation(Exp), Tautology(Taut)  I do not want to use conditional proof and indirect proof(reductio ad absurdum). Can I prove the above statement using only 19 rules?
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I'm taking a course from Stanford in Logic. I'm stuck with an exercise where I'm doing some proof. The Fitch system I'm given only allows   $ \land $ introduction and elimination $ \lor $ introduction and elimination $ \to $ introduction and elimination $ \equiv $ introduction and elimination $ \lnot$ introduction and  $ \lnot \lnot $ elimination   but not    $ \bot $ introduction and elimination   I've been struggling to prove the law of excluded middle (``p ∨ ¬p`) within this system. All of the proofs I've seen online make use of $\bot$ elimination to prove it by contradiction:  | ~(p | ~p)  (assumption) | | p        (assumption) | | p | ~p   (or introduction) | | $\bot$ | | | ~p       (assumption) | | p | ~p   (or introduction) | | $\bot$ | | (p | ~p)   (proof by contradiction)   Since I can't use $\bot$ elimination, I was trying to do this with $ \lnot$ introduction:  | ~(p | ~p)  (assumption) | | ??? | ~(p | ~p) =&gt; x | | | ??? | ~(p | ~p) =&gt; ~x | p | ~p     (not introduction)   but I just can't get it. I can't figure out how to get x and ~x.  Is this provable without $\bot$ elimination? If so, what are the steps?    Background:  What I'm ultimately trying to prove is : Given p ⇒ q, use the Fitch System to prove ¬p ∨ q. I'm able to prove that p ⇒ ¬p ∨ q and that ¬p ⇒ ¬p ∨ q, but I need to be able to plug p ∨ ¬p into the proof to be able to do 'or elimination'.
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Using only rules of natural deduction, I am trying to prove that $$p \implies q \vdash \lnot p \lor q$$ but am having a lot of difficulty. I was able to prove the other direction.   Could anyone show me or point me to the right direction?
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I have been attempting to use identities to get to the answer but I am unable to get anywhere. Here is the equation I am trying to prove tautological without using truth tables:  $[(p\rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r )$  And here is my attempt at using identities to work my way up. (a link to a page with a table of logical equivalence identities: )  $\lnot [(p \rightarrow q) \land (q \rightarrow r)] \lor (p \rightarrow r)$  $\lnot [(\lnot p \lor q) \land (q \rightarrow r)] \lor (p \rightarrow r)$  $[\lnot (\lnot p \lor q) \land \lnot (q \rightarrow r)] \lor (p \rightarrow r)$  $[(p\land \lnot q) \land (q \land \lnot r)] \lor (p \rightarrow r)$  I don't know where to go from there.
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Context  I'm having trouble understanding the limited situations in which qualifiers can be distributed.  I am given that the rules are:  $$\forall x\left[P(x)\land Q(x)\right]\equiv\forall xP(x) \land\forall xQ(x) \tag{1}$$ $$\exists x\left[P(x)\lor Q(x)\right]\equiv\exists xP(x) \lor\exists xQ(x) \tag{2}$$  $$\forall x\left[P(x)\lor Q(x)\right]\not\equiv\forall xP(x) \lor\forall xQ(x) \tag{3}$$ $$\exists x\left[P(x)\land Q(x)\right]\not\equiv\exists xP(x) \land\exists xQ(x) \tag{4}$$    Questions   What are examples of $P(x)$ and $Q(x)$ that illustrate when these statements are true? How can I show these statements always hold using proofs? Why can't "$\forall x$" be distributed in example 3 or "$\exists x$" in example 4?
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I came across the following proof in the book Logic, by Paul Tomassi:   (P & Q) → ~R : R → (P → ~Q)   According to the author, the proof should be a simple application of modus tollens. The following is perhaps the most obvious way of proving it:    1.   (P & Q) → ~R         Premise  2.   R                    Assumption  3.   ~~R                  Double Negative Ins.  4.   ~(P & Q)             1,3 Modus Tollens  5.   ~P ∨ ~Q              4 DeMorgen  6.   P → ~Q               5 Implication Ident.  7.   R → (P → ~Q)         2,6 Conditional Proof   The only problem is that the author hasn't yet introduced DeMorgen's identity or any other identities. The following is a list of all the resources that have so far been introduced:   Conjunction Introduction &amp; Elimination Conditional Introduction &amp; Elimination (Modus Ponens) Biconditional Introduction &amp; Elimination Double Negative Introduction &amp; Elimination Modus Tollens   However, I don't see any way to do the proof without identity rules. The following is a possible strategy for trying to solve it without identies:   1.   (P & Q) → ~R         Premise 2.   R                    Assumption 3.   ~~R                  Double Negative Ins. 4.   ~(P & Q)             1,3 Modus Tollens 5.   ~(P → ~Q)            Assumption ... 15.  ~(P → ~Q) → (P & Q)  5,14 Conditional Proof 16.  P → ~Q               4,15 Modus Tollens 17.  R → (P → ~Q)         2,16 Conditional Proof   But I don't see how it's possible to get to my theoretical step 15 without deriving the identies from scratch. According to the author, the proof is possible in 11 steps.  How can it be done?
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I am looking to for a way to prove that the statement $(p \vee q) \wedge (\neg p \vee r) \rightarrow (q \vee r)$ is a tautology. I am unable to use the truth table or the rules of inference for this. I can only solve this by using Laws and Theorems such as De Morgans Law and Distributive Law. Any help would be greatly appreciated. Thank you
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Okay I have the following fitch proof almost figured out, from here I know I have to move forth in conlcuding but I don't know what I am missing  (2 premises)  S → (R ∨ P)  P → (¬R → Q)   my work below  new scope assume S   Rule: > elim  R ∨ P  New scope assume R  Q∨R     Rule: V intro  End scope – (new scope assume P)  ¬R → Q         Rule: > Elim  my confusion comes where I go from here  Q∨R            rule Taut Con ??  Goal:├ S → (Q ∨ R)
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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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let D be a deductive system. The system's axiom set, A, includes all propositions that are not tautology, and the only rule of inference is $\frac{\alpha \vee \beta}{\alpha \wedge \beta}$. I need to answer:  Show with induction on the proof that $ \vdash \alpha \Longrightarrow \nvDash \alpha $, i.e if $ \alpha $ can be proven from the empty set, then exists a model $ M $ such that for every $ \varphi \in A $, $ M(\varphi)=T $ and $ M(\alpha)=F$. Is the system strongly sound? Is $X\nvDash \alpha \Longrightarrow X\vdash \alpha$? Prove that the empty set isn't consistent in D    I've already asked about this part of the question here but I still have a problem with it, because every axiom, $a\in A$,  $\vdash a$ and $A \models a$, in contradiction to what I'm supposed to prove here. Moreover because the axiom set includes all what isn't tautology it also includes all that is contradiction, which means that there is no model $ M $ such that for every $ \varphi \in A $, $ M(\varphi)=T $ which means for every proposition $\psi$, $A\Longrightarrow \psi$. This one was fairly easy base on (1) the system isn't sound so it can't be strongly sound. I'm not even sure how to approach this, I don't even have an intuition whether this is correct or not. Assuming this is wrong, I need to find a set $X$ and proposition $\alpha$ such that $X\nvDash \alpha$ but $\alpha$ cannot be proven from $X$, and if it's correct I need to show that for every set, $X$, and every proposition $\alpha$ such that $X\nvDash \alpha$, $\alpha$ can be proven from $X$ On the surface this looks pretty easy, both $(\varphi \wedge \psi)$ and $\neg(\varphi \wedge \psi)$ are axioms, as they aren't tautology, hence $\vdash (\varphi \wedge \psi)$ and $\vdash \neg (\varphi \wedge \psi)$ and the empty set isn't consistent but this seems too easy, like I'm missing something.   EDIT: I might have come up with a solution for (3), if $X\nvDash \alpha$ it means that $\alpha$ isn't a tautology and if it isn't a tautology it is an axiom in D and therefore can be proven from $X$
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I've scoured the math stackexchange and the philosophy one for some guidance on how to go about this while using the Fitch System.  Anyone can attempt it here;    I can reach the conclusion, but it's not in the correct "layer" of the proof. I'm stuck with the answer as a subproof.  (Note that the fitch system lacks contradiction elimination (RAA) as a method for getting out of assumptions).
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The fact that there exists irrational number $a,b$ such that $a^b$ is rational is proved by the law of excluded middle, but I read somewhere that irrationality of $\sqrt{2}^{\sqrt{2}}$ is proved constructively. Do you know the proof?
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Prove $ (A \cup B) \cap C$ =  $(A \cap C) \cup (B \cap C) $  Starting from the left side,   $ (A \cup B) \cap C = $  By distributive law, ( distributing the $\cap C$), we have  $ (A \cap C ) \cup (B \cap C) = $  Therefore,   $ (A \cap C ) \cup (B \cap C) = (A \cap C) \cup (B \cap C)$  If I start from the right, I have  =  $(A \cap C) \cup (B \cap C) $  By Distributive Law   =  $(A \cup B ) \cap C$  Therefore, $ (A \cup B) \cap C$ =  $(A \cup B ) \cap C$  Did I do this correctly or do I need to include the set union definition and the set intersection definitions?  Assuming that I need to include the set union definition of $A \cup B$ for the left side  $[x: x \in A \lor x \in B]$ so that means that x belongs in A or x belongs in B  For the right side I would have set intersection   $[x: x \in A \land x \in C]$ so x belongs in A and x belongs in C   $[x: x \in B \land x \in C]$ so x belongs in B and x belongs in C   so maybe it's like this?   $[x: x \in A \land x \in C] \lor [x: x \in B \land x \in C]$  and then by distributive law I would have gotten  $[x: x \in A \lor x \in B] \land C$  which becomes $(A \cup B) \cap C$  My question is how do I write a better proof than this jumbled mess?
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Here's another of Tomassi's exercises I can't solve (Logic, page 106):  ~ (~P &amp; ~Q) : P ∨ Q   I have to use natural deduction and the only rules I know are:   assumptions, modus ponendo ponens, modus tollendo tollens, double negation, reductio ad absurdum, conditional proof, v-introduction, v-elimination, and introduction, and elimination.   Tomassi's proof consists of 14 steps.  Thanks for the help.
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Context: I keep running into circular reasoning in attempting to derive strong induction (more generally "induction" whether it be weak or strong) from the well-ordering principle. Assume:     Peano Axioms w/ Well-Order: (I realize these are presented semi-informally)         (P1) $\ \exists x[x\!=\!1]$;   (P2) $\ $there exists an unary ("successor") function $S$ on $\mathbb{N}$ (i.e., $x\!\in\!\mathbb{N}\rightarrow S(x)\!\in\!\mathbb{N}$);   (P3) $\ \forall x[S(x)\!\neq\!1]$;   (P4) $\ \forall x\forall y[x\!\neq\!y\rightarrow S(x)\!\neq\!S(y)]$   (P5') $\ $Well-Ordering Principle: Every non-empty set includes a least-element: $$ \forall B\Big[\emptyset\!\subset\!B\!\subseteq\!\mathbb{N}\longrightarrow\exists m\big[m\!\in\!B\wedge\forall p[p\!\in\!B\rightarrow m\!\leq\!p]\big]\Big] $$   Note: Since I am trying to show the logical equivalency between Induction and the Well-Ordering Principle, I am assuming the latter as an axiom and treating the former as a theorem (below). The relation $\leq$ used above is defined next.         Order Relation         (O1) $\ x\!&lt;\!y\leftrightarrow\exists z[x\!+\!z\!=\!y]$   (O2) $\ x\!\leq\!y\leftrightarrow[x\!=\!y\vee x\!&lt;\!y]$         Addition (since order above is defined with "$+$")         (A1) $\forall x[S(x)=x+1]$   (A2) $\forall x\forall y[x+S(y)=S(x\!+\!y)]$      With this as background, below is the theorem and proof I see most often (or some variation thereof) in textbooks and online forums.     Theorem: The Well-Ordering Principle (P5') implies the Strong Induction Principle.      Proof: Suppose $X\!\subset\!\mathbb{N}$ with: (1) $1\!\in\!X$, and (2) $\forall x[x\!&lt;\!k\rightarrow x\!\in\!X]\rightarrow k\!\in\!X$. Assume $X'\!\equiv\!\mathbb{N}\setminus X$ is non-empty. From Well-Ordering Principle (P5'), $X'$ includes a least element $m$. From (1), $m\!\neq\!1$. Since $m$ is the least element of $X'$, then all $x\!&lt;\!m$ must belong to $X$. However, property (2) implies that $m\!\in\!X$, a contradiction. Thus, $X'$ must be empty and $X\!=\!\mathbb{N}$.   My problem is with the highlighted sentence above, which appears to me to require the result that $\neg(m\!\leq\!x)\leftrightarrow x\!&lt;\!m$, which I believe is based on the Trichotomy Property (exactly one of $x\!&lt;\!y$, $x\!=\!y$, $x\!&gt;\!y$ holds). However, the only proof of the Trichotomy Property (that I am aware of) based on the above axioms/definitions alone uses Induction (see ). Thus, it appears the reasoning is circular (induction is used to prove an intermediate result that is then used to prove induction).  Question: Assuming my understanding above is correct, my question simply put is this: does a proof exist of the Trichotomy Property for $&lt;$ as defined above that does not require a proof based on induction?
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I just started to study Mathematical Logic and I am having trouble finding a simple proof of the following exercise.    Let A be a propositional formula that contains only the $\equiv$ connective. For those who are not familiar with it $f_\equiv(T,T)=f_\equiv(F,F)=T$ and $f_\equiv(F,T)=f_\equiv(T,F)=F$. Show that A is a tautology if and only if each atomic proposition that appears in A appears an even number of times. I have managed to find a complicated proof using inductions and the fact that $\equiv$ is both commutative and associative but I don't believe this was the point of the exercise. Any help would be appreciated.
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I have to do a number of similar type questions, but I am having trouble grasping the general concepts around soundness and completeness. I have read up on the general definitions and think I understand them, but when asked to show something like the following I come up blank. Thanks for any ideas/help.   Show that the following propositional proof system, PCa, is sound but not complete.  PCa  Connectives (and constants) ∨, ∧, →, ↔, ¬, 0, 1  Axioms All tautologies of the form F→F  Rules of inference Hypothetical Syllogism
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I know how to prove this using an element and subset proof but that's not what I'm being asked to do. I'm suppose to prove it using the distributive law and the associative law but I'm just not seeing how to do this. Can you give me some advice on how to start this. I figure the first thing would be $$((A \cup B) \cap A) \cup ((A \cup B) \cap B^c))$$ But I'm not sure where to go from there.