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Proof step 8 from proof int2bits-leading-sgn (previous step)

0 ≤ i, ¬ 0 ≤ i ↔ uint2bits(i).rest.first ⊦ # uint2bits(i) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 ≤ i ⊦ # uint2bits(i) = 1, uint2bits(i).rest.first

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;