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Proof step 38 from proof bytes2bytes-zero (previous step)

8 = # fillsgn(8, int2bits(b2i(by0))), b2i(by) = 0, 7 < # fillsgn(8, int2bits(b2i(by0))), 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 + # fillsgn(8, int2bits(b2i(by0))) - 8, false ' + fillsgn(8, int2bits(b2i(by0))))) + i2b(bits2int(fillsgn(8, int2bits(b2i(by0))))) ' = by ' + by0 ', fillsgn(8, int2bits(b2i(by0))) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)