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Proof step 12 from proof add-eq2pot (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: (i2 + 1) * 2 ^ i0 - 2 ^ i0 = j

Additional information

Used simplifier rules: eq-two: ⊦ 0 ≤ i ∧ 0 ≤ j → (2 ^ i = 2 ^ j ↔ i = j);%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used formulas: (i2 + 1) * 2 ^ i0 - 2 ^ i0 = j