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Proof step 5 from proof restrict-add-in (
previous step
)
adr ∈ iast0, adr ∈ as, ∀ adr. adr ∈ iast0 | as → adr ∈ iast1 ∧ (iast0 | as)[adr] = iast1[adr] ⊦ # as = 0, as = ∅, adr ∈ iast1
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: adr
Additional information
Used simplifier rules: ⊦ adr ∈ iast | as ↔ adr ∈ iast ∧ adr ∈ as; (ws) Used terms: adr