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Proof step 7 from proof prefixn-size-prefixn (previous step)

# x < # (n ≤ # x ⊃ firstn(n, x);x) ⊦ n = 0, x = firstn(n, x)

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ n ≤ # x → # firstn(n, x) = n; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)