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Proof step 3 from proof disj-in-both (previous step)

n1 < m, n0 < m, disj(asar, 0, m), adr ∈ asar[n1], adr ∈ asar[n0] ⊦ # asar[n1] = 0, # asar[n0] = 0, asar[n1] = ∅, asar[n0] = ∅, n0 = n1

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:
n0 < m, n1 < m, adr ∈ asar[n0], adr ∈ asar[n1], ∀ n0, n1. n0 < m ∧ n1 < m ∧ n0 ≠ n1 → asar[0 + n0] ∩ asar[0 + n1] = ∅ ⊦ # asar[n0] = 0, # asar[n1] = 0, asar[n0] = ∅, asar[n1] = ∅, n0 = n1

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