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Proof step 41 from proof bytes2bytes-zero (previous step)

int2bits(b2i(by0)) ≠ [], 8 = # fillsgn(8, int2bits(b2i(by0))), 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(firstn(1, false ' + fillsgn(8, int2bits(b2i(by0))))) + i2b(bits2int(int2bits(b2i(by0)))) ' = by ' + by0 ', fillsgn(8, int2bits(b2i(by0))) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
8 = # fillsgn(8, int2bits(b2i(by0))), 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 ⊦ fillsgn(8, int2bits(b2i(by0))) = [], bits2bytes(firstn(1, false ' + fillsgn(8, int2bits(b2i(by0))))) = by '

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes1 + bytes0 ↔ bytes = bytes1;% ⊦ i2b(b2i(by)) = by;% ⊦ bits2int(int2bits(i)) = i;