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Proof step 3 from proof disj-write (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ disj(asar, n, m) ↔ (∀ n0, n1. n0 < m ∧ n1 < m ∧ n0 ≠ n1 → asar[n + n0] ∩ asar[n + n1] = ∅)
Proof step 5:
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 = ∅)

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, 0, asar] flags: <all paths>

Additional information

Used terms: m, 0, asar