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Proof step 133 from proof setInt-theorem (previous step)

¬ # int2bytes(i) ≥ 4, fillsgn(4, int2bytes(i)) = by ' + by0 ' + by1 ' + by2 ', # int2bytes(i) ≤ 4, by ∈ fillsgn(4, int2bytes(i)), by2 ∈ fillsgn(4, int2bytes(i)) ⊦ # int2bytes(i) = 3, # int2bytes(i) = 2, i = b2i(i2b(i)), by = i2b(i >> 24) ∧ by0 = i2b(i >> 16) ∧ by1 = i2b(i >> 8) ∧ by2 = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%ge: ⊦ n ≥ m ↔ m ≤ n; from specification nat