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

n0 + n1 ≤ # x ⊦ firstn(n0 + n1, x) = (n0 ≤ # x ⊃ firstn(n0, x);x) + (n1 ≤ # (n0 ≤ # x ⊃ restn(n0, x);[]) ⊃ firstn(n1, (n0 ≤ # x ⊃ restn(n0, x);[]));(n0 ≤ # x ⊃ restn(n0, x);[]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n0 + n1 ≤ # x ⊦ firstn(n0 + n1, x) = firstn(n0, x) + firstn(n1, restn(n0, x))

Rule argument

Additional information

Used simplifier rules: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)%n ≤ # x ⊦ # restn(n, x) = # x - n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)