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Proof step 9 from proof bib-neg-high (previous step)

¬(i < 0 ∧ lastn(8, int2bits(i)).first ∧ 8 ≤ # int2bits(i) ∧ 8 ≠ 0 ∧ ~ i ≠ 2 ^ n→i(8)), i < ~128, i ≤ ~256, ~128 ≤ i % 256, i < 0 ⊦ b2i(i2b(bits2int(lastn(8, int2bits(i))))) = i % 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%numlsledel: j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0); from specification int-abs