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Proof step 6 from proof restrict-add-subseteq-at (previous step)

adr ∈ iast0, adr ∈ iast0 | (as ++ adr) → adr ∈ iast1 ∧ (iast0 | (as ++ adr))[adr] = iast1[adr] ⊦ iast1[adr] = iast0[adr]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(iast0 | (as ++ adr))[adr] = iast1[adr], adr ∈ iast0, adr ∈ iast1 ⊦ iast1[adr] = iast0[adr]

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ as ++ adr0 ↔ adr = adr0 ∨ adr ∈ as; (s)% ⊦ adr ∈ iast | as ↔ adr ∈ iast ∧ adr ∈ as; (ws)