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Proof step 5 from proof disj-write (previous step)

n < # asar, n < m, m ≤ # asar, ∀ n0, n1. n0 < m ∧ n1 < m ∧ n0 ≠ n1 → asar[0 + n0] ∩ asar[0 + n1] = ∅ ⊦ disj(asar[n, as], 0, m) ↔ (∀ n0. n0 ≠ n ∧ n0 < m → asar[n0] ∩ as = ∅)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n < m, n < # asar, m ≤ # asar, ∀ n0, n1. n0 < m ∧ n1 < m ∧ n0 ≠ n1 → asar[n0] ∩ asar[n1] = ∅ ⊦ disj(asar[n, as], 0, m) ↔ (∀ n0. n0 ≠ n ∧ n0 < m → asar[n0] ∩ as = ∅)

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)