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Proof step 4 from proof restrict-further (previous step)

as1 ⊆ as0, iast0 | as0 ⊆ iast1 ⊦ ∀ adr. adr ∈ iast0 | as1 → adr ∈ iast1 ∧ (iast0 | as1)[adr] = iast1[adr]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
adr ∈ iast0, adr ∈ as1, adr ∈ as0, iast0 | as0 ⊆ iast1, as1 ⊆ as0 ⊦ # as1 = 0, # as0 = 0, iast0[adr] = iast1[adr], as0 = ∅, as1 = ∅

Rule argument

Additional information

Used simplifier rules: restrict-add-in: ⊦ iast0 | as ⊆ iast1 ∧ adr ∈ as ∧ adr ∈ iast0 → adr ∈ iast1;%size-zero-01: ⊦ as ≠ ∅ → # as ≠ 0; (forward) from specification set-basic[node-store]%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[node-store]%sub-in: ⊦ adr ∈ as ∧ as ⊆ as0 → adr ∈ as0; (forward) from specification set-basic[node-store]% ⊦ adr ∈ iast ∧ adr ∈ as → (iast | as)[adr] = iast[adr]; (s)% ⊦ adr ∈ iast | as ↔ adr ∈ iast ∧ adr ∈ as; (ws)