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Proof step 21 from proof size-eq-0 (previous step)

0 = as0.size, adr ∈ as0, ∀ adr. adr ∈ as0 → adr.size = 0, ∀ as1. # as1 < #(as0 ++ @(m, m0, 0)) → true → (0 = as1.size ↔ (∀ adr. adr ∈ as1 → adr.size = 0)) ⊦ # as0 = 0, adr.size = 0, as0 = ∅, @(m, m0, 0) ∈ as0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
0 = as0.size, adr = @(m1, m2, m3), adr ∈ as0, ∀ adr. adr ∈ as0 → adr.size = 0, ∀ as1. # as1 < #(as0 ++ @(m, m0, 0)) → true → (0 = as1.size ↔ (∀ adr. adr ∈ as1 → adr.size = 0)) ⊦ # as0 = 0, as0 = ∅, m3 = 0, @(m, m0, 0) ∈ as0

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, m1, m2, m3]

Additional information

Used simplifier rules: ⊦ (m = adr.lnum ∧ m0 = adr.pos) ∧ m1 = adr.size ↔ adr = @(m, m0, m1); (elim) Used terms: adr, m1, m2, m3