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Proof step 26 from proof bytes2bytes-zero (previous step)

fillsgn(8, int2bits(b2i(by0))) ≠ [] ∧ (fillsgn(8, int2bits(b2i(by0))).first ↔ ¬ false), b2i(by) = 0, b2i(by0) < 0, ∀ by0, by, bytes0. # bytes0 < # [] → b2i(by) = 0 ∧ b2i(by0) < 0 → bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0))) = by ' + by0 ' + bytes0 ⊦ bits2bytes(false ' + fillsgn(8, int2bits(b2i(by0)))) = by ' + by0 '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ j < 0 → int2bits(j).first;% ⊦ fillsgn(n, int2bits(j)).first ↔ int2bits(j).first;