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Proof step 8 from proof to-adrs-eq (previous step)

lax.to-adr(n) = lax0.to-adr(n), ∀ n, lax1. lax.to-adr(n) = lax1.to-adr(n) → lax = lax1, @(ladr.off, ladr.size) = ladr ∨ @(ladr.off, ladr.size) ∈ lax0 ⊦ @(ladr.off, ladr.size) ' + lax = ladr ' + lax0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
lax.to-adr(n) = lax0.to-adr(n), ladr = @(m, m0), ∀ n, lax1. lax.to-adr(n) = lax1.to-adr(n) → lax = lax1, @(m, m0) = ladr ∨ @(m, m0) ∈ lax0 ⊦ @(m, m0) ' + lax = ladr ' + lax0

Rule argument

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

Additional information

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