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Proof step 2 from proof fill-lastn-02 (previous step)

⊦ bits = [], lastn(m, fillsgn(m + n, bits)) = lastn(m, fillsgn(m, bits))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ bits ≠ [] ∧ m ≤ n → lastn(m, fillsgn(n, bits)) = lastn(m, fillsgn(m, bits))

Rule argument

name: fill-lastn-03 rotate: F subst: [bits, n, m] → [bits, m + n, m]

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ ¬ m + n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits, m + n, m