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Proof step 5 from proof prefixn-size-prefixn (previous step)

⊦ firstn(# (n ≤ # x ⊃ firstn(n, x);x), x) = (n ≤ # x ⊃ firstn(n, 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: ⊦ firstn(# x, x) = x; (s)% ⊦ n ≤ # x → # firstn(n, x) = n; (s)