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Could you please clarify, why this has been done here? 

A∃ ({p, q}, ¬q ∧ p, Φ1 ∧ Φ2, FG¬q ∨ FGp

A∃ ({q, p}, ¬q ∧ p, Φ1 ∧ Φ2, A∃ ({r}, ¬r, r0 ↔ (r ⇒ p | q), FG(¬r ∨ p)))

Can we do the operation like the below instead of the above one?

A∃ ({p, q}, ¬q ∧ p, Φ1 ∧ Φ2, FG[¬q ∨ p]

in # Study-Organisation (Master) by (1.1k points)

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FGφ ∨ FGψ is not equivalent to FG(φ ∨ ψ), but we can reduce the disjunction using a new state variable as shown on slide 51 of the related chapter. 

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FGϕ ∨ FGψ = A∃ ({q}, ¬q, q′ ↔ (q ⇒ ψ| ¬ϕ), FG[¬q ∨ ψ]) . Thanks for clarifying , Professor.
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