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I understand the solution but does my solution work too?

in * TF "Emb. Sys. and Rob." by (660 points)

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+1 vote

Let's check: 

  • EG!a holds in states {s1;s2;s3}
  • EF b holds in states {s2;s3}
  • E(F b & EG!a) is equivalent to EFb & EG!a which holds in states {s2;s3}
  • EF(b & EG!a) holds in states {s2;s3}

Hence, the two formulas S1 and S2 hold in exactly the same states of your structure, so that your structure does not distinguish between them.

by (166k points)

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