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Proof step 6 from proof uint2bits-truein (previous step)

0 ≤ j, 0 < j ⊦ j ≠ 0 ∧ ¬(∃ i. 0 ≤ i ∧ j = ~(2 ^ i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
2 ^ i + j = 0, 0 < j, 0 ≤ i ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ~ i = j ↔ i + j = 0;