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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# y = n0 + n1 ⊦ y = firstn(n0, y) + firstn(n1, restn(n0, 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)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ n ≤ # x → restn(n, x + y) = restn(n, x) + y; (s)