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Proof step 7 from proof fill-lastn-03 (previous step)

¬ n ≤ # bits, m < # bits, m ≤ n ⊦ lastn(m, fillsgn(n, bits)) = lastn(m, bits), bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m < n, # bits < n, m < # bits, m + 1 < n ⊦ bits = [], lastn(m, fillsgn(n, bits)) = lastn(m, bits)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n;% ⊦ m +1 = m + 1;% ⊦ m < n → n ≠ 0;% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;