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Proof step 65 from proof bib-neg-high (previous step)

8 = # fillsgn(8, int2bits(j0)), ~128 < j0, i0 * 256 + j0 < 0, ~128 ≤ j0, i0 * 256 + j0 ≤ ~256, i0 < 0, j0 < 0 ⊦ b2i(i2b(bits2int(fillsgn(8, int2bits(j0))))) = j0, ~256 ≠ i0 * 256 + j0 ∧ fillsgn(8, int2bits(j0)).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2bits-first-01: ⊦ j < 0 → int2bits(j).first; from specification int2bits%int2bits-fillfirst: ⊦ fillsgn(n, int2bits(j)).first ↔ int2bits(j).first; from specification int2bits%add-c: ⊦ i + j = j + i; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult