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Proof step 3 from proof prefix-prefixn-same (previous step)

⊦ (n ≤ # x ⊃ firstn(n, x);x) ⊑ x

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ x ⊑ x; (s)% ⊦ n ≤ # x → (firstn(n, x) ⊑ x ↔ true); (ss)