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Proof step 1 from proof cast-add-2pot-01 (conclusion)

⊦ 0 < i → cast(i, ~(j + 2 ^ i)) = cast(i, ~ j)

User applied rule cut formula

And obtained the following new goals

Proof step 2:
~(j + 2 ^ i) = ~1 * 2 ^ i + ~ j ⊦ 0 < i → cast(i, ~(j + 2 ^ i)) = cast(i, ~ j)

Rule argument

fma: ~(j + 2 ^ i) = ~1 * 2 ^ i + ~ j

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i * ~1 = ~ i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ ~ i + ~ j = ~(i + j);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~ i = ~ j ↔ i = j; Used formulas: ~(j + 2 ^ i) = ~1 * 2 ^ i + ~ j