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Proof step 17 from proof uint2bits-pred (previous step)

1 < j, 0 < j, 0 ≤ j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → uint2bits(j0 - 1) = sgn(bitsinvert(bitsinvert(uint2bits(j0)) +1)) ⊦ j % 2 ≠ 1, j = 2, uint2bits((j - 1) / 2) + ((j - 1) % 2 = 1)' = sgn(bitsinvert((bitsinvert(uint2bits(j / 2)) + (j % 2 ≠ 1)')+1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ bitsinvert(true ') = false ';% ⊦ bitsinvert(bitsinvert(bits)) = bits;% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + false ')+1 = bits + true;