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Proof step 10 from proof to-adrs-sublist (previous step)

1 ≤ n1, n1 ≤ # lax0 + 1, ∀ n0, n1, n. n0 + n1 ≤ # lax0 → sublist(n0, n1, lax0.to-adr(n)) = sublist(n0, n1, lax0).to-adr(n) ⊦ @(n, m, m0) ' + sublist(0, n1 - 1, lax0.to-adr(n)) = sublist(0, n1, @(m, m0) ' + lax0).to-adr(n), n1 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
n1 = m1 + 1, n1 ≤ # lax0 + 1, 1 ≤ n1, ∀ n0, n1, n. n0 + n1 ≤ # lax0 → sublist(n0, n1, lax0.to-adr(n)) = sublist(n0, n1, lax0).to-adr(n) ⊦ @(n, m, m0) ' + sublist(0, m1, lax0.to-adr(n)) = sublist(0, n1, @(m, m0) ' + lax0).to-adr(n), n1 = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n1, m1]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n1, m1