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Proof step 129 from proof initialization (previous step)

⌊err0⌋ ⊦ ⟨choose err0 with ⌊err0⌋ in err := err0⟩ (dirs1 = dirs1 ∧ files1 = files1 ∧ (boolvar ↔ boolvar) ∧ err0 = err)

Heuristic symbolic execution applied rule choose right

And obtained the following new goals

Rule argument

Additional information