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Proof step 2 from proof int2bits-truein (previous step)

⊦ j = 0, true ∈ int2bits(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i → int2bits(i) = uint2bits(i)
Proof step 4:
¬ 0 ≤ j ⊦ j = 0, true ∈ int2bits(j)

Rule argument

name: int2bits-pos rotate: F subst: [i] → [j]

Additional information

Used simplifier rules: uint2bits-onein: ⊦ 0 < j → true ∈ uint2bits(j);% ⊦ i ≠ j → (i ≤ j ↔ i < j); Used terms: j