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Proof step 5 from proof mod-two-ge-01 (previous step)

j < i, 0 < i, 0 ≤ j ⊦ 2 ^ i = 2 ^ j * ((2 ^ i - 1) / 2 ^ j + 1)

User applied rule cut formula

And obtained the following new goals

Proof step 6:
2 ^ i = 2 ^ j * 2 ^ (i - j), 0 < i, j < i, 0 ≤ j ⊦ 2 ^ i = 2 ^ j * ((2 ^ i - 1) / 2 ^ j + 1)
Proof step 13:
0 < i, j < i, 0 ≤ j ⊦ 2 ^ i = 2 ^ j * ((2 ^ i - 1) / 2 ^ j + 1), 2 ^ i = 2 ^ j * 2 ^ (i - j)

Rule argument

fma: 2 ^ i = 2 ^ j * 2 ^ (i - j)

Additional information

Used formulas: 2 ^ i = 2 ^ j * 2 ^ (i - j)