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Proof step 6 from proof prefixn-toset-restr-empty (previous step)

ax.toSet | m = ∅ ⊦ #(firstn(n, ax).toSet | m) = 0, firstn(n, ax).toSet | m = ∅, (n ≤ # ax ⊃ firstn(n, ax);ax).toSet | m = ∅

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
ax.toSet | m = ∅, n ≤ # ax ⊦ #(firstn(n, ax).toSet | m) = 0, firstn(n, ax).toSet | m = ∅

Rule argument

pos: right 3

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