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Proof step 3 from proof prefixn-toset-inter-empty (previous step)

ax.toSet ∩ as = ∅ ⊦ #(prefixn(n, ax).toSet ∩ as) = 0, prefixn(n, ax).toSet ∩ as = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ prefixn(n, ax) = (n ≤ # ax ⊃ firstn(n, ax);ax)
Proof step 5:
ax.toSet ∩ as = ∅ ⊦ #((n ≤ # ax ⊃ firstn(n, ax);ax).toSet ∩ as) = 0, (n ≤ # ax ⊃ firstn(n, ax);ax).toSet ∩ as = ∅

Rule argument

spec: list-prefixn, inst: address-list-set name: prefixn seq: ⊦ prefixn(n, ax) = (n ≤ # ax ⊃ firstn(n, ax);ax) subst: [ax, n] → [ax, n] flags: <all paths>

Additional information

Used terms: ax, n