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Proof step 6 from proof prefixn-prefixn (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
(m ≤ # x ⊃ firstn(m, x);x) = y + z, (# y < m ⊃ # y;m) ≤ # x ⊦ y = firstn((# y < m ⊃ # y;m), x)
Proof step 15:
(m ≤ # x ⊃ firstn(m, x);x) = y + z ⊦ y = x, (# y < m ⊃ # y;m) ≤ # x

Rule argument

pos: right 1

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