Modus Tollens on {B|α} → R?

[entropy1]

I can't figure out this: Say we have event B given α, denoted as {B|α}. If B happens, that implies that R happens: {B|α} → R.

Now I want to apply modus Tollens. So if I do, do I get the result: ¬R → {¬B|α}? I mean, I hope I can keep the α unaffected. Is that the case? ¬X meaning X does not happen.

 

[Mark44]

I can't figure out this: Say we have event B given α, denoted as {B|α}. If B happens, that implies that R happens: {B|α} → R.

This doesn't seem right to me. What you have written looks like a conditional probability, rather than one of the usual operations used in symbolic logic; e.g. conjunction, disjunction, negation, or implication. For starters B and A (let's use A rather than α) are both events, each with its own probability of occurring. The conditional probability Pr(B | A) is defined as:
$$Pr(B | A) = \frac{Pr(A \wedge B)}{Pr(A)}$$
With symbolic logic an expression is either true or false, so probability doesn't enter into the calculations. It seems to me you are mixing symbolic logic and probability.

Now I want to apply modus Tollens. So if I do, do I get the result: ¬R → {¬B|α}? I mean, I hope I can keep the α unaffected. Is that the case? ¬X meaning X does not happen.

 

[entropy1]

I forgot to mention that the α signifies all other factors are left unchanged. Maybe that is reasonable? My original formulation was {B, α}, a loose formulation.

 

[fresh_42]

I do not see how $\alpha$ is necessary here. Just put it into $B'$. So if $B' \to R$ and $\lnot R$ then modus tollendo tollens says $\lnot B'$. Now that we have $B'=\{B\,|\,\alpha\}$ as given $\alpha$, you can independently investigate $\{B\,|\,\alpha\}$ and $\{B\,|\,\lnot \alpha\}.$

 

[entropy1]

B is a specific event out of set {}B. ¬B is any event different from event B out of the same set. Likewise for R.

Suppose that {B, α} → R (1) and {¬B, α} → ¬R (2), where α is an element out of the set of possible circumstances for events out of set {}B.

Then (2) would be the same as {B, α} ← R, would it? R implies B, but not the circumstances α. α is more or less a given.

 

[fresh_42]

I am officially confused by your special anysets of circumstances. Reminds me a bit of the famous anykey. My first reaction was: draw me a Venn diagram.

You can only conclude $R\longrightarrow \lnot \{\lnot B,\alpha\}$. Whether $\lnot \{\lnot B,\alpha\} \longleftrightarrow \{B,\alpha\}$ cannot be said from the information you gave us.

 

[Jarvis323]

I can't figure out this: Say we have event B given α, denoted as {B|α}. If B happens, that implies that R happens: {B|α} → R.

Now I want to apply modus Tollens. So if I do, do I get the result: ¬R → {¬B|α}? I mean, I hope I can keep the α unaffected. Is that the case? ¬X meaning X does not happen.

It sounds like you're just using {B|a} to mean B and a. The negation would be not R implies (not B or not a).

 

[entropy1]

It sounds like you're just using {B|a} to mean B and a. The negation would be not R implies not B or not a.

Exactly what I was thinking.

It seems to come down to (Bn AND α) → Rn, or ¬Rn → (¬Bn OR ¬α).

Thanks!

 

[Mark44]

Exactly what I was thinking.

It seems to come down to (Bn AND α) → Rn, or ¬Rn → (¬Bn OR ¬α).

Thanks!

That makes more sense. Here you're using one of de Morgan's Laws to convert $\neg (B_n \wedge \alpha)$ to $(\neg B_n \vee \neg \alpha)$.

 

[sysprog]

The modus tollendo tollens rule is a logic rule that is usually called modus tollens and which can be derived via transposition from the modus ponens rule − some writers refer to modus tollendo ponens and some writers refer to modus ponendo tollens -- anyway the modus ponens rule is that if A then B, and A, then B, and the modus tollens rule is that if A then B, and not B, then not A.