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Proof step 11 from proof in-restr-notin-log (previous step)

adr ∈ as, adr ∈ abs-log(nx, gstore), adr ∈ ax ⊦ #(ax.toSet ∩ abs-log(nx, gstore).toSet) = 0, # as = 0, as = ∅, adr.lnum ∈ nx

Heuristic elimination applied rule insert elim lemma

And closed the goal!

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] → [adr, m, m0, m1]

Additional information

Used simplifier rules: abs-log-adr-explicit-in-leb-notin: ⊦ ¬ n ∈ nx → ¬ @(n, n0, n1) ∈ abs-log(nx, gstore);% ⊦ (m = adr.lnum ∧ m0 = adr.pos) ∧ m1 = adr.size ↔ adr = @(m, m0, m1); (elim) Used terms: adr, m, m0, m1