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Is there a way to simplify this sort of statement? For example, $$a \Rightarrow (b\Rightarrow c)$$ is equivalent to $$(a \wedge b) \Rightarrow c.$$ I'm looking for something similar for $$(a\Rightarrow b) \Rightarrow (p \Rightarrow q),$$ if it's even possible.

Willemien
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Spine Feast
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3 Answers3

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Hint: $$ (a \Rightarrow b) \Leftrightarrow ((\neg a) \lor b)$$

Edit: One way to continue from $(a \wedge \neg b ) \vee \neg p \vee q$ would be: $$ (a \land \neg b ) \lor \neg p \lor q $$ $$ \lnot(\lnot((a \land \neg b ) \lor \neg p)) \lor q $$ $$ \lnot(\lnot(a \land \neg b ) \land p) \lor q $$ $$ \lnot((\lnot a \lor b ) \land p) \lor q $$ $$ \lnot((a \Rightarrow b) \land p) \lor q $$ $$ ((a \Rightarrow b) \land p) \Rightarrow q $$

Vincent Pfenninger
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  • Using this, I got $(a \wedge \neg b ) \vee \neg p \vee q$. Not sure what to do with this now. – Spine Feast Jul 23 '13 at 12:33
  • @DepeHb see my Edit – Vincent Pfenninger Jul 23 '13 at 12:48
  • Ok. I guess there would be a simpler final form but it seems that the best one can do is $ ((a \Rightarrow b) \land p) \Rightarrow q $. So this could have been gotten exactly from $x \Rightarrow (y \Rightarrow z) \iff (x \wedge y)\Rightarrow z$ by letting $x = (a \Rightarrow b), \ y=p, \ z=q$ – Spine Feast Jul 23 '13 at 13:39
  • It is not quite clear to me what you mean by simpler form. What sort of form would you consider to be simple? – Vincent Pfenninger Jul 23 '13 at 14:10
  • I guess something with only one implication sign. I tried to get it to that form via a truth table, considering $a \wedge b \wedge p \Rightarrow q$ and found out the two only differ at two out of sixteen possible values for the variables, so in principle the two cases could be added to one side to yield an equivalence to a statement with only one implication, but it's a fairly messy one. – Spine Feast Jul 23 '13 at 14:13
  • Well you can always get rid of an implication using the identity I first mentioned. – Vincent Pfenninger Jul 23 '13 at 14:24
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Apply the point @Vincent noted, for all statements in $(a\to b)\to(p\to q)$. Then use the following fact for some proper statements $A$, $B$ and $C$: $$(A\to B)\vee(C\to B)\equiv (A\wedge C)\to B$$

Mikasa
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Type theory implies the following are equivalent:

  • $(a \to b) \to (p \to q)$,
  • $((a \to b) \land p) \to q$,
  • $p \to ((a \to b) \to q)$.

I hope this helps ;-)

dtldarek
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