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I have the following statement that I want to prove:$(a → (b → c)) ∧ (∼ c) ≡ (a → ∼ b) ∧ (∼ c)$

I think I can prove this using the law of equivalences, however I also noticed that both statements, the LHS and the RHS has a ∧ (∼ c) at the end.

So is it fine that I conclude $(a → (b → c)) = (a → ∼ b)$?

Can someone please explain why or why not I can do this? I think I have done this before in normal algebra but not with statement variables.

Probability_Sarah
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  • You need to prove $(b → c)∧(∼ c) = ∼ b$. – JMP Mar 17 '23 at 06:42
  • @Milten Yes I know. I had tried to put a tilde in there but didn't know the mathjax. Will erase comment and try again. – coffeemath Mar 17 '23 at 07:27
  • Generally $(p \to q) = (\lnot p∨q).$ – coffeemath Mar 17 '23 at 07:29
  • this is not true. let $ \phi $ be $c \land \lnot c$ then for every $\alpha, \beta$ its obviously true that $\alpha \land \phi = \beta \land \phi$. by the logic of your qeustion $\alpha = \beta$ the difference from normal algebra is the lack of inverse here. if $a + c = b +c $ then $a + c - c = b + c - c$. – yairb Mar 17 '23 at 07:49

3 Answers3

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I want to prove: $$(a → (b → c)) ∧ (\lnot c) \quad≡\quad (a → \lnot b) ∧ (\lnot c)$$

It is invalid to just "cancel out" the $(\land \lnot c):$ observe that if $a,b$ and $c$ each means that T. rex is extinct, doing so results in the false biconditional $$(a → (b → c)) \quad↔\quad (a → ¬b).$$

In general, $$A↔B \quad\models\quad A∧C↔B∧C,$$ while $$A∧C↔B∧C \quad\not\models\quad A↔B;$$ the latter is because any inequivalence of $A$ and $B$ is "disrupted" whenever a false $C$ is conjuncted with each.

P.S. $P\models Q$ means that no matter what meanings I assign to sentences $P$ and $Q,\:P$ implies $Q.$ On the other hand, $P\not\models Q$ means that it is possible to assign meanings such that sentence $P$ is true and sentence $Q$ false.

ryang
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  • thanks ryang but what are the two weird symbols? – Probability_Sarah Mar 18 '23 at 15:36
  • @SarahV.P Thanks for asking! I just added a postscript; $⊨$ means "logically implies", but as we're in a purely abstract propositional-logic setting, you can rewrite the two symbols as "$⇒$" and "$\not\Rightarrow$" and read them as "implies" and "does not imply", respectively. – ryang Mar 18 '23 at 17:25
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Close but not quite.

Either $c$ is true or false.

In the case where $c$ is true, then the statements are both false (as $X\land\lnot\top$ is false whatever $X$ may be).

In the case where $c$ is false, then you should establish whether $(a\to(b\to \bot))$ is equivalent to $a\to\lnot b$.

Graham Kemp
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  • Thanks Graham, but can you explain your symbols you used? like the T and the upside down T? – Probability_Sarah Mar 18 '23 at 15:43
  • @s $⊤$ is a sentence that is true regardless of what the sentence names ($A,B,P,X,$ etc.) in the rest of the formalisation stand for; it is called a tautology. On the other hand, $⊥$ is a sentence that is false...; it's called a contradiction. $\quad$ The atomic sentence P is neither, since it can variously represent "T. rex is extinct" and "1=3". – ryang Mar 18 '23 at 17:44
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Your Core Query is Basically this :

Is it Correct/valid to "Prove" $X \land Z \equiv Y \land Z$ by Proving $X \equiv Y$ , ignoring the Common $Z$ ?

Answer : No

Here are 2-3 ways to look at it , to get Intuition , which will then let us analyse your Specific Case at the End :

(A) In terms of Equations with Integers :
Proving $ax=bx$ (where $a$ & $b$ are Arbitrarily Complicated Expressions) may be done with Proving $a=b$ , but these two may not be Equal in the given Scenario: It may be that $x=0$ & we can never Prove that $a=b$ , though we can Prove the overall $ax=bx$

(B) In terms of Matrix Equations :
We may be given to Prove $AC=BC$ , but that gives $(A-B)C=0$ where we can neither conclude that $C=0$ nor conclude that $A-B=0$ (which is $A=B$) : Here too , we can not ignore the Common $C$ , & we have to show the overall Equality.

(C) In terms of Mathematical Logic :
In $X \land Z \equiv Y \land Z$ , where each term is some logical Expression with some "logical variables" , it might be that :
(C1) With certain values of the "logical variables" , $X \equiv Y$ & overall , it is true
(C2) With certain other values of those "logical variables" , $X \not \equiv Y$ , but in those cases $Z=0$ , hence overall , it is true.

In that Scenario , we may never be able to Prove $X \equiv Y$ (because it is not true) , though $X \land Z \equiv Y \land Z$ is true.

(D) In terms of your given Example :
It is not true. When $\lnot c$ is true , it becomes $0 \equiv 0$ , which is Entirely true , Independent of $a$ & $b$ !
In that Case , we can see that your Simplified Expression Depends on $a$ & $b$ !
Hence , these 2 can not be logically Equivalent.

Prem
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