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Proof step 1 from proof int2bits-truein-01 (conclusion)

⊦ true ∈ int2bits(j).rest ↔ j ≠ 0 ∧ ¬(∃ i. 0 ≤ i ∧ j = ~(2 ^ i))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
true ∈ int2bits(j).rest ⊦ j ≠ 0 ∧ ¬(∃ i. 0 ≤ i ∧ j = ~(2 ^ i))
Proof step 4:
j ≠ 0 ∧ ¬(∃ i. 0 ≤ i ∧ j = ~(2 ^ i)) ⊦ true ∈ int2bits(j).rest

Rule argument

pos: right 1

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