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Proof step 5 from proof in-prefixn-in-shorter-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ firstn(n0, x), a ∈ x, n0 ≤ n1, n1 ≤ # x, n0 ≤ # x ⊦ a ∈ firstn(n1, x)

Rule argument

Additional information

Used simplifier rules: firstN-in-04: ⊦ a ∈ firstn(n, x) ∧ n ≤ # x → a ∈ x; (forward) from specification list-del%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat