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Proof step 40 from proof bib-neg-high (previous step)

0 < 1 - 2 * i0 ∧ 1 - 2 * i0 ≠ 1, i0 < 0 ⊦ b2i(i2b(bits2int(sgn(bitsinvert(uint2bits(~ i0)) + true ').last ' + false ' + false ' + false ' + false ' + false ' + false ' + false '))) = ~128, i0 = 0, i0 * 256 ≠ ~128 ∧ sgn(bitsinvert(uint2bits((1 - 2 * i0) / 2) + ((1 - 2 * i0) % 2 = 1)) +1).last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: succ-zero: ⊦ (bits + false ')+1 = bits + true; from specification bitopsaux%a: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; from specification list[bitlista]%inv-one-01: ⊦ bitsinvert(true ') = false '; from specification bitopsaux%inv-append: ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0); from specification bitopsaux%Last: ⊦ bits + boolvar = bits + boolvar '; from specification list[bitlista]%mod-two-pred-02: ⊦ i < 0 → (1 - 2 * i) % 2 = 1; from specification int-div%div2-odd-01: ⊦ i < 0 → (1 - i * 2) / 2 = ~ i; from specification int-div%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%nummult-ls: ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i); from specification int-div%ax-04: ⊦ i + 0 = i; from specification int-basic2%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; 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%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%two-zero: ⊦ 2 * i = 0 ↔ i = 0; from specification int-mult%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs