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Proof step 7 from proof disj-rec (previous step)

n0 < m, n < m, ∀ n0, n1. n0 < m + 1 ∧ n1 < m + 1 ∧ n0 ≠ n1 → asar[0 + n0] ∩ asar[0 + n1] = ∅ ⊦ #(asar[n0] ∩ asar[n]) = 0, n0 = n, asar[n0] ∩ asar[n] = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n < m, n0 < m, ∀ n1, n2. ¬ m < n1 ∧ ¬ m < n2 ∧ n1 ≠ n2 → asar[n1] ∩ asar[n2] = ∅ ⊦ #(asar[n0] ∩ asar[n]) = 0, n0 = n, asar[n0] ∩ asar[n] = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)