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

n0 + n1 ≤ # lax0 + 1, ∀ n0, n1, n. n0 + n1 ≤ # lax0 → sublist(n0, n1, lax0.to-adr(n)) = sublist(n0, n1, lax0).to-adr(n) ⊦ sublist(n0, n1, @(n, ladr.off, ladr.size) ' + lax0.to-adr(n)) = sublist(n0, n1, ladr ' + lax0).to-adr(n)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
ladr = @(m, m0), n0 + n1 ≤ # lax0 + 1, ∀ n0, n1, n. n0 + n1 ≤ # lax0 → sublist(n0, n1, lax0.to-adr(n)) = sublist(n0, n1, lax0).to-adr(n) ⊦ sublist(n0, n1, @(n, m, m0) ' + lax0.to-adr(n)) = sublist(n0, n1, ladr ' + lax0).to-adr(n)

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