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Proof step 9 from proof uint2bits-first (previous step)

uint2bits(i / 2) ≠ [], 0 ≤ i, 0 < i, uint2bits(i / 2).first, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ uint2bits(i0).first → false ⊦ i = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 ≤ i, 0 < i, uint2bits(i / 2).first, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ uint2bits(i0).first → false ⊦ i = 1

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;