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Proof step 10 from proof prefixn-append-slice (previous step)

# y ≤ n0 ⊦ y + (n0 ≤ # z + # y ⊃ firstn(n0 - # y, z);z) = (n0 ≤ # y + # z ⊃ firstn(n0, y + z);y + z)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 11:
# y ≤ n0, n0 ≤ # z + # y ⊦ y + firstn(n0 - # y, z) = (n0 ≤ # y + # z ⊃ firstn(n0, y + z);y + z)

Rule argument

pos: right 1

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic% ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list