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Proof step 18 from proof bytes2bytes-zero (previous step)

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(sgn(fillsgn(8, int2bits(b2i(by))) + byte2bits(by0))) = by ' + by0 '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
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(sgn(false ' + byte2bits(by0))) = by ' + by0 '

Rule argument

Additional information

Used simplifier rules: ⊦ sgn(boolvar ' + boolvar ' + bits) = sgn(boolvar ' + bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ mklist(boolvar, 0) = [];%n ≠ 0 ⊦ mklist(false, (* n)) = false ' + mklist(false, n -1);% ⊦ 0 < n → fillsgn(n, boolvar ') = mklist(boolvar, n);% ⊦ int2bits(0) = false ';