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I need to prove:

$$p\rightarrow (\Box (\Box p \wedge p) \rightarrow (\Box p \wedge p))$$

The system contains all propostional tautologies and the axiom scheme $\mathbf K$:$ \Box(p \rightarrow q) \rightarrow (\Box p \rightarrow \Box q) $.

Rules are modus ponens, substitution and necessitation.

Thanks for help!

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  • $\begingroup$ What does $\Box$ stand for? Negation? $\endgroup$ – layman Nov 26 '14 at 12:58
  • $\begingroup$ @MathIsHardNoItsNot Necessity. $\endgroup$ – Git Gud Nov 26 '14 at 13:00
  • $\begingroup$ @Charles Do you mind listing $\bf K$? $\endgroup$ – Git Gud Nov 26 '14 at 13:00
  • $\begingroup$ @GitGud Sure, no problem! $\endgroup$ – Charles Bronson Nov 26 '14 at 13:02
  • $\begingroup$ Do you have the reflexivity axiom $\square p\to p$? $\endgroup$ – user21467 Nov 26 '14 at 13:14
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$\square p\land p\to p$ is a theorem; therefore $\square(\square p\land p\to p)$ by necessitation, and so $$ \square(\square p\land p)\to \square p $$ by $\mathbf K$ (and modus ponens). Now apply $p\land\cdot$ to both sides and use the equivalence of $a\land b\to c$ and $a\to(b\to c)$.

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  • $\begingroup$ great job! Thank you! $\endgroup$ – Charles Bronson Nov 26 '14 at 13:37

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