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Proof step 1 from proof fill-lastn-04 (conclusion)

⊦ m + 1 = n ∧ # bits0 < n → lastn(m, bits + fillsgn(n, bits0).rest) = fillsgn(n -1, bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# bits0 ≤ m ⊦ lastn(m, bits + fillsgn(m + 1, bits0).rest) = fillsgn(m, bits0)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ (m + 1)-1 = m;