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Proof step 4 from proof ladrlist-to-adr-split (previous step)

n < # lax ⊦ lax.to-adr(m)[n] = @(m0, m1, m2) ↔ (∃ lax0, lax1. lax = lax0 + @(m1, m2) ' + lax1 ∧ m0 = m ∧ # lax0 = n), lax = []

User applied rule structural induction

And obtained the following new goals

Proof step 5:
n < #(ladr + lax0), ∀ n, m, m0, m1, m2. n < # lax0 → (lax0.to-adr(m)[n] = @(m0, m1, m2) ↔ (∃ lax2, lax1. lax0 = lax2 + @(m1, m2) ' + lax1 ∧ m0 = m ∧ # lax2 = n)) ∨ lax0 = [] ⊦ ladr + lax0 = [], (ladr + lax0).to-adr(m)[n] = @(m0, m1, m2) ↔ (∃ lax2, lax1. ladr + lax0 = lax2 + @(m1, m2) ' + lax1 ∧ m0 = m ∧ # lax2 = n)

Rule argument

indvar: lax

Additional information

Used simplifier rules: laddress-list freely generated by [], +;;