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Proof step 4 from proof fill-lastn-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m < # bits, m ≤ n ⊦ lastn(m, fillsgn(n, bits)) = lastn(m, bits), bits = []

Rule argument

Additional information

Used simplifier rules: fill-lastn: ⊦ n < # bits → lastn(n, fillsgn(n, bits)) = lastn(n, bits);% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;