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Proof step 113 from proof setInt-theorem (previous step)

0 ≤ b2i(i2b(i >> 16)), 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)

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 3 flags: discard

Additional information

Used simplifier rules: ibi: ⊦ i2b(b2i(by)) = by; from specification byte%shiftright-byte-8-pos: ⊦ 0 ≤ b2i(by) → b2i(by) >> 8 = 0; from specification byte%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs