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Proof step 1 from proof to-adrs-sublist (conclusion)

n0 + n1 ≤ # lax ⊦ sublist(n0, n1, lax.to-adr(n)) = sublist(n0, n1, lax).to-adr(n)

User applied rule structural induction

And obtained the following new goals

Proof step 2:
n0 + n1 ≤ #(ladr + lax0), ∀ n0, n1, n. n0 + n1 ≤ # lax0 → sublist(n0, n1, lax0.to-adr(n)) = sublist(n0, n1, lax0).to-adr(n) ⊦ sublist(n0, n1, (ladr + lax0).to-adr(n)) = sublist(n0, n1, ladr + lax0).to-adr(n)

Rule argument

indvar: lax

Additional information

Used simplifier rules: adrs-empty: ⊦ ([]).to-adr(n) = [];%laddress-list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ sublist(0, 0, ax) = []; (s)% ⊦ sublist(0, 0, lax) = []; (s)