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Proof step 4 from proof firstn-last-eq (previous step)

# x = n0 + n, n ≤ # x ⊦ firstn(n, x) + lastn(n0, x) = x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# x = n0 + n ⊦ firstn(n, x) + lastn(n0, x) = x

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)