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Proof step 71 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(i % 2 ^ n→i(8))) = i % 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 72:
i < ~256, 8 ≤ # int2bits(i), lastn(8, int2bits(i)).first, i < 0, ~128 ≤ i % 256 ⊦ ~256 = i, b2i(i2b(i % 256)) = i % 256

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ []; from specification int2bits%l: ⊦ # bits = 0 ↔ bits = []; from specification list-data[bitlista]%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%numlsledel: j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0); from specification int-abs%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