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Proof step 6 from proof to-adrs-sublist (previous step)

n0 = 0, 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, @(m, m0) ' + lax0).to-adr(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n1 ≤ # lax0 + 1, ∀ n0, n1, n. n0 + n1 ≤ # lax0 → sublist(n0, n1, lax0.to-adr(n)) = sublist(n0, n1, lax0).to-adr(n) ⊦ sublist(0, n1, @(n, m, m0) ' + lax0.to-adr(n)) = sublist(0, n1, @(m, m0) ' + lax0).to-adr(n)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)