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Proof step 11 from proof prefixn-split-sublist (previous step)

# y = n0 + n1, n0 + n1 ≤ # y ⊦ y = firstn(n0, y) + sublist(n0, n1, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# y = n0 + n1 ⊦ y = firstn(n0, y) + sublist(n0, n1, y)

Rule argument

Additional information

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