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Proof step 5 from proof uint2bits-falsein-01 (previous step)

0 < j ∧ j ≠ 1, 0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ false ∈ (uint2bits(j / 2) + (j % 2 = 1)).rest ↔ (∀ i. 2 ^ i ≠ j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ j = 1, false ∈ (uint2bits(j / 2) + (j % 2 = 1)').rest ↔ (∀ i. 2 ^ i ≠ j + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar = bits + boolvar ';