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Proof step 1 from proof int2bits-mod256 (conclusion)

⊦ 0 < j ∧ 0 ≤ j0 ∧ j0 < 256 → int2bits(j * 256 + j0) = int2bits(j) + fillsgn(9, int2bits(j0)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < j, 0 ≤ j0, j0 < 256 ⊦ int2bits(j * 256 + j0) = int2bits(j) + fillsgn(9, int2bits(j0)).rest

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)