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Proof step 9 from proof prefixn-prefixn (previous step)

# y0 = (# y < m ⊃ # y;m), (m ≤ # y0 + # z0 ⊃ firstn(m, y0 + z0);y0 + z0) = y + z ⊦ y = y0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 10:
# y0 = (# y < m ⊃ # y;m), firstn(m, y0 + z0) = y + z, m ≤ # y0 + # z0 ⊦ y = y0
Proof step 13:
# y0 = (# y < m ⊃ # y;m), y0 + z0 = y + z ⊦ y = y0, m ≤ # y0 + # z0

Rule argument

pos: left 2

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