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

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

User applied rule insert equation

And obtained the following new goals

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

Rule argument

pos: left 1 flags: discard

Additional information