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Proof step 3 from proof to-adrs-in (previous step)

∀ m, m0, m1, n. @(m, m0, m1) ∈ lax0.to-adr(n) ↔ m = n ∧ @(m0, m1) ∈ lax0 ⊦ @(m, m0, m1) = @(n, ladr.off, ladr.size) ∨ @(m, m0, m1) ∈ lax0.to-adr(n) ↔ m = n ∧ (@(m0, m1) = ladr ∨ @(m0, m1) ∈ lax0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
ladr = @(m2, m3), ∀ m, m0, m1, n. @(m, m0, m1) ∈ lax0.to-adr(n) ↔ m = n ∧ @(m0, m1) ∈ lax0 ⊦ @(m, m0, m1) = @(n, m2, m3) ∨ @(m, m0, m1) ∈ lax0.to-adr(n) ↔ m = n ∧ (@(m0, m1) = ladr ∨ @(m0, m1) ∈ lax0)

Rule argument

spec: laddress name: elim seq: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0) subst: [ladr, m, m0] → [ladr, m, m2]

Additional information

Used simplifier rules: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0); (elim) Used terms: ladr, m, m2