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Proof step 5 from proof add-eq2pot (previous step)

2 ^ i1 = i2 + 1, 0 ≤ j, 0 ≤ i0, 0 ≤ i1 + i0 ⊦ 2 ^ (i1 + i0) = 2 ^ i0 + j ↔ j = 2 ^ i0 * i2 ∧ 0 ≤ i1

User applied rule case distinction

And obtained the following new goals

Proof step 6:
2 ^ (i1 + i0) = 2 ^ i0 + j, 2 ^ i1 = i2 + 1, 0 ≤ j, 0 ≤ i0, 0 ≤ i1 + i0 ⊦ j = 2 ^ i0 * i2 ∧ 0 ≤ i1
Proof step 13:
j = 2 ^ i0 * i2 ∧ 0 ≤ i1, 2 ^ i1 = i2 + 1, 0 ≤ j, 0 ≤ i0, 0 ≤ i1 + i0 ⊦ 2 ^ (i1 + i0) = 2 ^ i0 + j

Rule argument

pos: right 1

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