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Proof step 4 from proof fill-lastn-04 (previous step)

¬ m ≤ # fillsgn(m + 1, bits0).rest, # bits0 ≤ m ⊦ lastn(m, bits + fillsgn(m + 1, bits0).rest) = fillsgn(m, bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# fillsgn(m + 1, bits0).rest < m, # bits0 ≤ m ⊦ lastn(m, bits + fillsgn(m + 1, bits0).rest) = fillsgn(m, bits0)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;