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Proof step 61 from proof setInt-theorem (previous step)

i >> 8 = b2i(i2b(i >> 8)), 0 ≤ i, 0 < i, 128 ≤ i, 0 ≤ i >> 8 ⊦ i = b2i(i2b(i)), i2b(0) = i2b(i >> 24) ∧ i2b(0) = i2b(i >> 16)

User applied rule cut formula

And obtained the following new goals

Proof step 62:
i >> 16 = (i >> 8) >> 8 ∧ i >> 24 = (i >> 8) >> 16, i >> 8 = b2i(i2b(i >> 8)), 0 ≤ i, 0 < i, 128 ≤ i, 0 ≤ i >> 8 ⊦ i = b2i(i2b(i)), i2b(0) = i2b(i >> 24) ∧ i2b(0) = i2b(i >> 16)
Proof step 68:
i >> 8 = b2i(i2b(i >> 8)), 0 ≤ i, 0 < i, 128 ≤ i, 0 ≤ i >> 8 ⊦ i >> 16 = (i >> 8) >> 8 ∧ i >> 24 = (i >> 8) >> 16, i = b2i(i2b(i)), i2b(0) = i2b(i >> 24) ∧ i2b(0) = i2b(i >> 16)

Rule argument

fma: i >> 16 = (i >> 8) >> 8 ∧ i >> 24 = (i >> 8) >> 16

Additional information

Used formulas: i >> 16 = (i >> 8) >> 8 ∧ i >> 24 = (i >> 8) >> 16