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Proof step 6 from proof lebs-del-subseteq-set (previous step)

adr0 ∈ as, as.lebs ⊆ ns ⊦ # as = 0, adr0 = adr, as = ∅, adr0.lnum ∈ ns, (as -- adr).lebs ⊆ as.lebs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
adr0 = @(m, m0, m1), adr0 ∈ as, as.lebs ⊆ ns ⊦ # as = 0, adr0 = adr, as = ∅, m ∈ ns, (as -- adr).lebs ⊆ as.lebs

Rule argument

spec: address-data name: elim seq: ⊦ (m = adr.lnum ∧ m0 = adr.pos) ∧ m1 = adr.size ↔ adr = @(m, m0, m1) subst: [adr, m, m0, m1] → [adr0, m, m0, m1]

Additional information

Used simplifier rules: ⊦ (m = adr.lnum ∧ m0 = adr.pos) ∧ m1 = adr.size ↔ adr = @(m, m0, m1); (elim) Used terms: adr0, m, m0, m1