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Proof step 9 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, uint2bits((j - 1) / 2) + ((j - 1) % 2 = 1)' = sgn(bitsinvert((bitsinvert(uint2bits(j / 2)) + (j % 2 ≠ 1)')+1))

User applied rule cut formula

And obtained the following new goals

Proof step 10:
j % 2 ≠ 1, 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, uint2bits((j - 1) / 2) + ((j - 1) % 2 = 1)' = sgn(bitsinvert((bitsinvert(uint2bits(j / 2)) + (j % 2 ≠ 1)')+1))
Proof step 17:
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))

Rule argument

fma: j % 2 ≠ 1

Additional information

Used formulas: j % 2 ≠ 1