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Proof step 5 from proof prefixn-eq-append (previous step)

x = (n ≤ # x + # y ⊃ firstn(n, x + y);x + y), # x < n ⊦ y = []

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
x = firstn(n, x + y), # x < n, n ≤ # x + # y ⊦ y = []

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ x = x + y ↔ y = []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list