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Proof step 6 from proof bits2int-zero-ls (previous step)

byte2bits(by) + bytes2bits(bytes0) ≠ [], ¬ 0 ≤ bits2int(bytes2bits(bytes0)) ↔ b2i(bytes0.first) < 0 ⊦ (byte2bits(by) + bytes2bits(bytes0)).first ↔ ¬ 0 ≤ b2i(by), bytes0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
¬ 0 ≤ bits2int(bytes2bits(bytes0)) ↔ b2i(bytes0.first) < 0 ⊦ byte2bits(by) + bytes2bits(bytes0) = [], bytes0 = [], (byte2bits(by) + bytes2bits(bytes0)).first ↔ ¬ 0 ≤ b2i(by)

Proof step 6 from proof bits2int-zero-ls (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information