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Proof step 14 from proof prefixn-split-dropn (previous step)

⊦ 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);[])), n0 + n1 ≤ # x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# x < n0 + n1 ⊦ x = (n0 ≤ # x ⊃ firstn(n0, x);x) + (n1 ≤ # (n0 ≤ # x ⊃ restn(n0, x);[]) ⊃ firstn(n1, []);(n0 ≤ # x ⊃ restn(n0, x);[]))

Rule argument

Additional information

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