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Proof step 111 from proof setInt-theorem (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 112:
i >> 24 = (i >> 16) >> 8, i >> 16 = b2i(i2b(i >> 16)), 0 < i >> 8, 128 ≤ i >> 8, 128 ≤ i, 0 ≤ i >> 8, 0 < i, 0 ≤ i ⊦ i = b2i(i2b(i)), i2b(0) = i2b(i >> 24)
Proof step 114:
i >> 16 = b2i(i2b(i >> 16)), 0 < i >> 8, 128 ≤ i >> 8, 128 ≤ i, 0 ≤ i >> 8, 0 < i, 0 ≤ i ⊦ i >> 24 = (i >> 16) >> 8, i = b2i(i2b(i)), i2b(0) = i2b(i >> 24)

Rule argument

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

Additional information

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