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Proof step 29 from proof bytes2bytes-zero (previous step)

¬ 8 < #(false ' + 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(false ' + 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 + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ m < n ↔ n ≤ m;%n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;