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

∀ n, m. dropn(n, lax0.to-adr(m)) = dropn(n, lax0).to-adr(m) ⊦ dropn(n, @(m, ladr.off, ladr.size) ' + lax0.to-adr(m)) = dropn(n, ladr ' + lax0).to-adr(m)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
ladr = @(m0, m1), ∀ n, m. dropn(n, lax0.to-adr(m)) = dropn(n, lax0).to-adr(m) ⊦ dropn(n, @(m, m0, m1) ' + lax0.to-adr(m)) = dropn(n, ladr ' + lax0).to-adr(m)

Rule argument

spec: laddress name: elim seq: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0) subst: [ladr, m, m0] → [ladr, m0, m1]

Additional information

Used simplifier rules: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0); (elim) Used terms: ladr, m0, m1