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Proof step 5 from proof prefixn-split-sublist (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# y = n0 + n1 ⊦ y = firstn(n0, y + z) + sublist(n0, n1, y + z)

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 + n; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)