Skip to main content
MathOverflow

Return to Answer

deleted 20 characters in body
Source Link
Bjørn Kjos-Hanssen
Bjørn Kjos-Hanssen
  • 24.8k
  • 3
  • 58
  • 114

I'll write $\to$ instead of $\supset$, and $\bot$ instead of false, below. Since Law of Excluded Middle is given, I'll argue using classical propositional logic.

Since $M\to\bigcirc M$ is already given as Axiom $\bigcirc$R, let's prove $$\bigcirc M\to M.$$ We are given $$\neg\bigcirc\bot.\tag{*}$$ We need to showFirst, by Axiom $\bigcirc R$, $$\neg M\to\neg\bigcirc M,$$$$\neg M\wedge\bigcirc M\to\bigcirc \neg M\wedge\bigcirc M.$$ in other wordsTherefore by $\bigcirc S$, $$A\to\neg\bigcirc\neg A.$$$$\neg M\wedge\bigcirc M\to\bigcirc (\neg M\wedge M)$$ So it suffices to showTherefore by definition of $\bot$, $$A\wedge\bigcirc\neg A\to\bot$$$$\neg M\wedge\bigcirc M\to\bigcirc\bot$$ so byBy (*) it suffices to obtain, $$A\wedge\bigcirc\neg A\to \bigcirc\bot.$$$$\neg(\neg M\wedge\bigcirc M)$$ But we have by Axiom $\bigcirc R$By de Morgan and Axiom $\bigcirc S$law of excluded middle, $$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc\bot$$$$M\vee \neg\bigcirc M$$ so we're done.So, $$\bigcirc M\to M$$

Note that Axiom $\bigcirc M$ was not needed.

I'll write $\to$ instead of $\supset$, and $\bot$ instead of false, below. Since Law of Excluded Middle is given, I'll argue using classical propositional logic.

Since $M\to\bigcirc M$ is already given as Axiom $\bigcirc$R, let's prove $$\bigcirc M\to M.$$ We are given $$\neg\bigcirc\bot.\tag{*}$$ We need to show $$\neg M\to\neg\bigcirc M,$$ in other words $$A\to\neg\bigcirc\neg A.$$ So it suffices to show $$A\wedge\bigcirc\neg A\to\bot$$ so by (*) it suffices to obtain $$A\wedge\bigcirc\neg A\to \bigcirc\bot.$$ But we have by Axiom $\bigcirc R$ and Axiom $\bigcirc S$, $$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc\bot$$ so we're done.

Note that Axiom $\bigcirc M$ was not needed.

I'll write $\to$ instead of $\supset$, and $\bot$ instead of false, below. Since Law of Excluded Middle is given, I'll argue using classical propositional logic.

Since $M\to\bigcirc M$ is already given as Axiom $\bigcirc$R, let's prove $$\bigcirc M\to M.$$ We are given $$\neg\bigcirc\bot.\tag{*}$$ First, by Axiom $\bigcirc R$, $$\neg M\wedge\bigcirc M\to\bigcirc \neg M\wedge\bigcirc M.$$ Therefore by $\bigcirc S$, $$\neg M\wedge\bigcirc M\to\bigcirc (\neg M\wedge M)$$ Therefore by definition of $\bot$, $$\neg M\wedge\bigcirc M\to\bigcirc\bot$$ By (*), $$\neg(\neg M\wedge\bigcirc M)$$ By de Morgan and law of excluded middle, $$M\vee \neg\bigcirc M$$ So, $$\bigcirc M\to M$$

Note that Axiom $\bigcirc M$ was not needed.

deleted 46 characters in body
Source Link
Bjørn Kjos-Hanssen
Bjørn Kjos-Hanssen
  • 24.8k
  • 3
  • 58
  • 114

I'll write $\to$ instead of $\supset$, and $\bot$ instead of false, below. Since Law of Excluded Middle is given, I'll argue using classical propositional logic.

Since $M\to\bigcirc M$ is already given as Axiom $\bigcirc$R, let's prove $$\bigcirc M\to M.$$ We are given $$\neg\bigcirc\bot\tag{*}$$ $$\bigcirc\bigcirc M\to\bigcirc M\tag{$\dagger$}$$ $$\bigcirc M\wedge\bigcirc N\to\bigcirc (M\wedge N)$$$$\neg\bigcirc\bot.\tag{*}$$ We need to show $$\neg M\to\neg\bigcirc M.$$$$\neg M\to\neg\bigcirc M,$$ Inin other words $$A\to\neg\bigcirc\neg A.$$ So it suffices to show $$A\wedge\bigcirc\neg A\to\bot$$ so by (*) it suffices to obtain $$A\wedge\bigcirc\neg A\to \bigcirc\bot$$$$A\wedge\bigcirc\neg A\to \bigcirc\bot.$$ But we have by Axiom $\bigcirc R$ and Axiom $\bigcirc S$, $$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc(\bot)$$$$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc\bot$$ so we're done. I guess I didn't use ($\dagger$).

Note that Axiom $\bigcirc M$ was not needed.

I'll write $\to$ instead of $\supset$, and $\bot$ instead of false, below. Since Law of Excluded Middle is given, I'll argue using classical propositional logic.

Since $M\to\bigcirc M$ is already given as Axiom $\bigcirc$R, let's prove $$\bigcirc M\to M.$$ We are given $$\neg\bigcirc\bot\tag{*}$$ $$\bigcirc\bigcirc M\to\bigcirc M\tag{$\dagger$}$$ $$\bigcirc M\wedge\bigcirc N\to\bigcirc (M\wedge N)$$ We need to show $$\neg M\to\neg\bigcirc M.$$ In other words $$A\to\neg\bigcirc\neg A.$$ So it suffices to show $$A\wedge\bigcirc\neg A\to\bot$$ so by (*) it suffices to obtain $$A\wedge\bigcirc\neg A\to \bigcirc\bot$$ But we have $$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc(\bot)$$ so we're done. I guess I didn't use ($\dagger$).

I'll write $\to$ instead of $\supset$, and $\bot$ instead of false, below. Since Law of Excluded Middle is given, I'll argue using classical propositional logic.

Since $M\to\bigcirc M$ is already given as Axiom $\bigcirc$R, let's prove $$\bigcirc M\to M.$$ We are given $$\neg\bigcirc\bot.\tag{*}$$ We need to show $$\neg M\to\neg\bigcirc M,$$ in other words $$A\to\neg\bigcirc\neg A.$$ So it suffices to show $$A\wedge\bigcirc\neg A\to\bot$$ so by (*) it suffices to obtain $$A\wedge\bigcirc\neg A\to \bigcirc\bot.$$ But we have by Axiom $\bigcirc R$ and Axiom $\bigcirc S$, $$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc\bot$$ so we're done.

Note that Axiom $\bigcirc M$ was not needed.

Source Link
Bjørn Kjos-Hanssen
Bjørn Kjos-Hanssen
  • 24.8k
  • 3
  • 58
  • 114

I'll write $\to$ instead of $\supset$, and $\bot$ instead of false, below. Since Law of Excluded Middle is given, I'll argue using classical propositional logic.

Since $M\to\bigcirc M$ is already given as Axiom $\bigcirc$R, let's prove $$\bigcirc M\to M.$$ We are given $$\neg\bigcirc\bot\tag{*}$$ $$\bigcirc\bigcirc M\to\bigcirc M\tag{$\dagger$}$$ $$\bigcirc M\wedge\bigcirc N\to\bigcirc (M\wedge N)$$ We need to show $$\neg M\to\neg\bigcirc M.$$ In other words $$A\to\neg\bigcirc\neg A.$$ So it suffices to show $$A\wedge\bigcirc\neg A\to\bot$$ so by (*) it suffices to obtain $$A\wedge\bigcirc\neg A\to \bigcirc\bot$$ But we have $$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc(\bot)$$ so we're done. I guess I didn't use ($\dagger$).