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Proof step 57 from proof bytes2int-append (previous step)

0 ≤ bits2int(bytes2bits(bytes0)) ∧ 0 ≤ 8 * n→i(# bytes0), 0 ≤ 2 ∧ 0 ≤ 8 ∧ 0 ≤ n→i(# bytes0), bytes2bits(bytes).first ⊦ bytes = [], bytes0 = [], bytes2bits(bytes) + bytes2bits(bytes0) = [], bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)) = [], bits2int(bytes2bits(bytes0)) % 2 ^ (8 * n→i(# bytes0)) + 2 ^ (8 * n→i(# bytes0)) = 2 ^ (8 * n→i(# bytes0)) + bits2int(bytes2bits(bytes0)), bytes2bits(bytes0).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
bytes2bits(bytes).first, 0 ≤ bits2int(bytes2bits(bytes0)) ⊦ bytes = [], bytes0 = [], bytes2bits(bytes) + bytes2bits(bytes0) = [], bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)) = [], bits2int(bytes2bits(bytes0)) % 2 ^ (8 * n→i(# bytes0)) = bits2int(bytes2bits(bytes0))

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)%bits ≠ [] ⊦ 0 ≤ bits2int(bits) → ¬ bits.first;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ n→i(n);% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = i + i0 ↔ j = i0;% ⊦ i * j = j * i;