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Proof step 3 from proof mod-twoodd-03 (previous step)

2 * i + 1 = ~(2 * (~ i - 1) + 1), i < j, i < 0, 0 < j, i + 1 < j ⊦ (i + 1) % j * 2 = (2 * i + 1) % (2 * j) + 1

User applied rule insert equation

And obtained the following new goals

Proof step 4:
i < j, i < 0, 0 < j, i + 1 < j ⊦ (i + 1) % j * 2 = ~(2 * (~ i - 1) + 1) % (2 * j) + 1

Rule argument

pos: left 1 flags: discard

Additional information