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Proof step 11 from proof prefixn-append-slice (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# y ≤ n0, n0 ≤ # z + # y ⊦ y + firstn(n0 - # y, z) = firstn(n0, y + z)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic