Prove that p $\rightarrow$(q$\rightarrow$p) is logically equivalent to $\neg p$ $\rightarrow$(p$\rightarrow$q) without using truth table. It is easy to show that both the statements are tautologies. Can we prove the result directly?
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4What rules do you have available? – Git Gud Jun 23 '15 at 18:04
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2Every tautology is provable by completeness; thus, having proved e.g. $\vdash \lnot p \to (p \to q)$ you can add to the proof an "extra premise" to derive e.g. : $p \to (q \to p) \vdash \lnot p \to (p \to q)$. – Mauro ALLEGRANZA Jun 23 '15 at 18:10
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Do I have to use Wajsberg's 1931 axiom DDpDqrDDDsrDDpsDpsDpDpq? – Doug Spoonwood Jun 23 '15 at 19:02
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3It is quite easy to derive $\vdash [p \to (q \to p)] \leftrightarrow [¬p \to (p \to q)]$ with Natural Deduction. – Mauro ALLEGRANZA Jun 23 '15 at 19:45
3 Answers
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Use that $a\rightarrow b$ is equivalent to (or, in fact, defined to be) $(\neg a)\vee b$ and write out both sides.
Damian Reding
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I think it's actually simpler to avoid that approach for this question. – Doug Spoonwood Jun 23 '15 at 19:47
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$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\then}{\rightarrow} \newcommand{\followsfrom}{\leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} $Writing $\;\equiv\;$ for logical equivalence, the simplest approach is to use the rule that $$ \tag{0} A \then B \;\equiv\; \lnot A \lor B $$ With this, the first expression can be rewritten as follows: $$\calc p \then (q \then p) \op\equiv\hints{the above rule $\ref 0$, twice} \hint{-- we can leave out the parentheses since $\;\lor\;$ is associative} \lnot p \lor \lnot q \lor p \op\equiv\hints{symmetry, i.e., $\;A \lor B \;\equiv\; B \lor A\;$;}\hint{excluded middle, i.e., $\;\lnot A \lor A \;\equiv\; \true\;$} \true \lor \lnot q \op\equiv\hint{$\;\true \lor A \;\equiv\; \true\;$} \true \endcalc$$
In other words, the first expression is indeed a tautology.
Now do something similar with the second expression, et voilà.
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I don't agree that this is simplest approach. I also can't get to your conclusion from what you've written. You started with (in Polish notation) CpCqp. Then you derived ANpANqp. That involves a certain complication and won't work in a definition-free system, but it is legal in some systems. Then you told me you use Aab == Aba and ANaa to get to A1Nq. But, starting from ANpANqp I can only get to ANpApNq using Aab == Aba. Thus, I'm going to downvote this as incorrect. – Doug Spoonwood Jun 23 '15 at 19:36
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@DougSpoonwood I explicitly mentioned that $;\lor;$ is associative ($;AaAbc \equiv AAabc;$), which gets us from $;ANpApNq;$ to $;AANppNq;$, from which we get $;A1Nq;$ and therefore $;1;$. I see nothing incorrect. And I admit that simplicity lies in the eye of the beholder: if --like me, and possibly like the OP-- one is very familiar with calculational proofs (see, e.g., EWD1300) and very unfamiliar with Polish notation, then your answer is not simple at all. But I'm not going to downvote it for that... – MarnixKlooster ReinstateMonica Jun 23 '15 at 19:51
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Your explanation amounts to saying that ¬p∨¬q∨p stands for two distinct well-formed formulas. Consequently, it is NOT clear what you did with your first step when you went from p→(q→p) to ¬p∨¬q∨p you went from CpCqp to ANpANqp or from CpCqp to AANpNqp, without the proof analysis. A proof consists of a unique sequence of well-formed formulas. You have not suggested such a unique sequence here, because you've used ¬p∨¬q∨p to stand for two different formulas simultaneously. Logical equivalence does not imply that you can notate two different wffs simultaneously and correctly in a proof. – Doug Spoonwood Jun 23 '15 at 21:55
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$\lnot p \lor \lnot q \lor p$ stands for two distinct equivalent well-formed formulas. So choosing one interpretation over the other forces a choice that limits our manipulation choices. I suggest you read EWD1300 to see my point of view. Basically, we're using a different notion of 'proof' here. The usefulness of our answers determines on the notion used by the OP, and by their teacher/textbook's. (I won't go further into that discussion here: I'm not good at doing that clearly. I've been here before, as you can see at another answer of mine: http://math.stackexchange.com/a/752877/11994.) – MarnixKlooster ReinstateMonica Jun 24 '15 at 04:41
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"I suggest you read EWD1300 to see my point of view. " First off what he writes up until the part about infix notation actually supports my view. It isn't clear how to check your argument. You can't leave out parentheses when checking an argument for a propositional calculus with equality and some equality axioms assumed. He suggests teaching proof design, but you certainly aren't doing that by writing ¬p∨¬q∨p, because that doesn't appear in any proofs in propositional calculus... only in proof sketches at best. And why can you just drop parentheses if an operation is associative? – Doug Spoonwood Jun 24 '15 at 07:46
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As far as I can tell, that is a rabbit pulled out of a magician's hat! Associativity by itself even does NOT guarantee that you can drop parentheses. It isn't the case that A $\lor$ B $\land$ C is clear, even though $\lor$ and $\land$ both associate. Then the author writes something like this "One would like the little kids to spend their time on learning the techniques of effective thought, but regrettably they spend a lot of it on familiarizing themselves with odd, irregular notational conventions whose wide adoption is their only virtue." But infix notation and it's conventions for... – Doug Spoonwood Jun 24 '15 at 07:59
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the precedence of operations, such as PEDMAS, or whatever it's called consists of exactly such a thing. I mean, sure exponentiation results in larger numbers faster, but addition and subtraction come as much simpler in terms of mathematical logic and you can build a lot from addition. Lastly what you wrote above does NOT consist of small, explicit steps. Even not taking into account that you condensed translation of CpCqp to it's AN form in one step, in the very same step you tried to go from ANpANqp to AANppq. That combines an association with a commutation, and is NOT small. – Doug Spoonwood Jun 24 '15 at 08:07
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I use Lukasiewicz/Polish notation.
So, we know that CpCqp is a tautology. Thus, CpCqp is logically equivalent to CpCNqp, since it qualifies as a special case of CpCqp (substitute 'q' with 'Nq' in CpCqp). One equivalence says that CNxy == CNyx, where x and y are wffs. Thus, since CpCNqp is logically equivalent to CpCNpq. The law of commutation says that CxCyz == CyCxz. Thus, CpCNpq is logically equivalent to CNpCpq. Or in a different format:
axiom 1 CpCqp
special case of 1 2 CpCNqp
CNxy == CNyx, 2 3 CpCNpq
CxCyz == CyCxz 4 CNpCpq
The implicit meta-lemma which makes this reversible goes that if x and y are tautologies, then any given particular formula x is logically equivalent to any of it's generalizations y (this doesn't imply that all tautologies are logically equivalent so far as I can tell). In other words, if x and y are tautologies, then if we can obtain x from y just by substitution, then x and y are logically equivalent. Thus we can write:
axiom 1 CNpCpq
CxCyz == CyCxz 2 CpCNpq
CNxy == CNyx 3 CpCNqp
3 is a special case of CpCqp
4 CpCqp
Doug Spoonwood
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