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Proof step 4 from proof fill-lastn-05 (previous step)

bits0 ≠ [] ∧ # bits0 < m + 1, # bits0 ≤ m ⊦ lastn(m, (bits0.first ' + fillsgn((m + 1)-1, bits0)).rest) = fillsgn(m, bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# bits0 ≤ m ⊦ bits0 = [], lastn(m, fillsgn(m, bits0)) = fillsgn(m, bits0)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ (m + 1)-1 = m;