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Proof step 15 from proof prefixn-eq-prefixn-app (previous step)

z ≠ [] ∧ # y = n1 ∨ z = [] ∧ # y ≤ n1 ∧ x1 = [] ∨ z = [] ∧ n1 = # y ∧ x1 ≠ [] ⊦ y = (n1 ≤ (# y + # z) + # x1 ⊃ firstn(n1, (y + z) + x1);(y + z) + x1)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 16:
n1 ≤ (# y + # z) + # x1, z ≠ [] ∧ # y = n1 ∨ z = [] ∧ # y ≤ n1 ∧ x1 = [] ∨ z = [] ∧ n1 = # y ∧ x1 ≠ [] ⊦ y = firstn(n1, (y + z) + x1)
Proof step 17:
z ≠ [] ∧ # y = n1 ∨ z = [] ∧ # y ≤ n1 ∧ x1 = [] ∨ z = [] ∧ n1 = # y ∧ x1 ≠ [] ⊦ y = (y + z) + x1, n1 ≤ (# y + # z) + # x1

Rule argument

pos: right 1

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