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

n < m, n0 < m, disj(asar, 0, m + 1) ⊦ #(asar[n0] ∩ asar[n]) = 0, n0 = n, asar[n0] ∩ asar[n] = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ disj(asar, n, m) ↔ (∀ n0, n1. n0 < m ∧ n1 < m ∧ n0 ≠ n1 → asar[n + n0] ∩ asar[n + n1] = ∅)
Proof step 7:
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] = ∅

Rule argument

spec: <toplevel> name: disj seq: ⊦ disj(asar, n, m) ↔ (∀ n0, n1. n0 < m ∧ n1 < m ∧ n0 ≠ n1 → asar[n + n0] ∩ asar[n + n1] = ∅) subst: [m, n, asar] → [m + 1, 0, asar] flags: <all paths>

Additional information

Used terms: m + 1, 0, asar