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Proof step 112 from proof setInt-theorem (previous step)

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 cut formula

And obtained the following new goals

Proof step 113:
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)

Rule argument

fma: 0 ≤ b2i(i2b(i >> 16))

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%shiftright-not-neg: ⊦ 0 ≤ i → ¬ i >> j < 0; from specification intops Used formulas: 0 ≤ b2i(i2b(i >> 16))