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Proof step 8 from proof uint2bits-second (previous step)

uint2bits(i / 2) ≠ [], 0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → uint2bits(i0).rest.first ⊦ i = 1, (uint2bits(i / 2).rest + (i % 2 = 1)').first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → uint2bits(i0).rest.first ⊦ i = 1, (uint2bits(i / 2).rest + (i % 2 = 1)').first

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;