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Proof step 15 from proof uint2bits-falsein-01 (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 16:
false ∈ uint2bits(j / 2).rest ↔ (∀ i. 2 ^ i ≠ j / 2 + 1), 0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ j % 2 = 0, j = 1, j % 2 ≠ 1 ∨ false ∈ uint2bits(j / 2).rest ↔ (∀ i. 2 ^ i ≠ j + 1)

Rule argument

subst: [j0] → [j / 2]

Additional information

Used simplifier rules: ⊦ 0 < j → (j < 2 ↔ j = 1);% ⊦ 0 < j → (i / j ≤ 0 ↔ i < j);% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);%j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j);% ⊦ 0 < j → abs(j) = j;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < i → ¬ i ≤ i / 2; Used terms: j / 2