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Proof step 64 from proof setInt-theorem (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
i >> 16 = 0, i >> 24 = b2i(i2b(i >> 8)) >> 16, 0 ≤ b2i(i2b(i >> 8)), 0 ≤ i, 0 < i, 128 ≤ i ⊦ i = b2i(i2b(i)), i2b(0) = i2b(i >> 24)

Rule argument

Additional information

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