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Proof step 2 from proof restrict-add-subseteq-at (previous step)

iast0[adr] = ind, adr ∈ iast0, iast0 | (as ++ adr) ⊆ iast1 ⊦ iast1[adr] = iast0[adr]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information