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Can some of you nice people help me and show me how to prove $\diamond \Box A$ → A characterizes symmetry.

I really appreciate it Bests

Rob Arthan
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Norman
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  • What have you tried? Are you sure you have got the statement of the problem right? – Rob Arthan Dec 09 '18 at 00:22
  • Can you spell out what does it mean to 'characterize symmetry' here? One direction of the statement is straightforward.. – Berci Dec 09 '18 at 01:23
  • @RobArthan, all i know is we have to define a Frame F = <W,R> and some world like s,t in W and if sRt then tRs and prove in both direction. but don't know how – Norman Dec 09 '18 at 11:48
  • @Berci i think this means that in this modal system if we define 2 world w,t in W and have a frame F<W,R> if sRt then tRs. but i don't know how should i prove it. – Norman Dec 09 '18 at 11:51
  • Ok. One statement is that if $R$ is symmetric in a frame, then all Kripke models on this frame satisfy the given formula. (This is the straightforward direction.) The other statement to prove is: if all Kripke models on a fixed frame satisfy the given formula, then the frame is symmetric. Can you prove any of these statements? – Berci Dec 09 '18 at 12:06
  • @Berci thank you for your reply. but no my problem was i don't know how should i prove this statement. i knew what should i do but i don't know how. – Norman Dec 09 '18 at 15:13
  • If $W = {a, b, c}$ and the accessibility relation is $R = {(a, b), (b, c), (c, a)}$, then $\Diamond \Box A \to A$ is valid under any valuation, but $R$ is not symmetric. – Rob Arthan Dec 09 '18 at 22:01
  • No. Let $A$ hold only at $c$, then $b\Vdash\Box A$ so $a\Vdash\diamond\Box A$ but $a\not\Vdash A$. – Berci Dec 09 '18 at 22:15

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Hints:

  1. If $(W, R)$ is a frame with $R$ symmetric, assume $(W, R, p), v\Vdash \diamond\Box A$ and deduce that $A$ is valid at $v$.
  2. If $(W,R)$ is not symmetric, there are $v, w\in W$ such that $vRw$ but not $wRv$, then define a valuation $p$ (of $A$) that makes $v\Vdash \diamond\Box A$ but $A$ is false at $v$.
Berci
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