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Proof step 130 from proof initialization (previous step)

⌊err1⌋ ⊦ ∃ err0. ⌊err0⌋ ∧ ⟨err := err0⟩ (dirs1 = dirs1 ∧ files1 = files1 ∧ (boolvar ↔ boolvar) ∧ err1 = err)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 131:
⌊err1⌋ ⊦ ∃ err0. ⌊err0⌋ ∧ ⟨err := err0⟩ (err1 = err)

Rule argument

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