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Proof step 3 from proof band-011x-pos-02 (previous step)

0 ≤ j ∧ ¬ false ∈ int2bits(i).rest ∧ 0 ≤ i ⊦ 0 ≤ j ∧ ¬ false ∈ int2bits(i).rest ∧ 0 ≤ i → j % 2 ^ n→i(# int2bits(i) - 1) = j % (i + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 ≤ j, 0 ≤ i ⊦ j % 2 ^ n→i(# int2bits(i) - 1) = j % (i + 1), false ∈ int2bits(i).rest

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)