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Proof step 4 from proof in-prefixn-in-shorter-prefix (previous step)

a ∈ (n0 ≤ # x ⊃ firstn(n0, x);x), n0 ≤ n1 ⊦ a ∈ (n1 ≤ # x ⊃ firstn(n1, x);x)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
a ∈ (n0 ≤ # x ⊃ firstn(n0, x);x), n0 ≤ n1, n1 ≤ # x ⊦ a ∈ firstn(n1, x)
Proof step 10:
a ∈ (n0 ≤ # x ⊃ firstn(n0, x);x), n0 ≤ n1 ⊦ a ∈ x, n1 ≤ # x

Rule argument

pos: right 1

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