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Proof step 67 from proof initialization (previous step)

⌊err1⌋ ⊦ ∃ err0. ⌊err0⌋ ∧ ⟨err := err0⟩ (err1 = err), ⟨err := err1⟩ (err1 = err)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

pos: right 2

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