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Proof step 94 from proof bytes2int-append (previous step)

bytes2bits(bytes0).first ∧ bytes2bits(bytes0) ≠ [], 0 ≤ 2 ^ (8 * n→i(# bytes0)) + ~ bits2uint(bitsinvert(bytes2bits(bytes0)) +1), ~ bits2uint(bitsinvert(bytes2bits(bytes0)) +1) < 0 ⊦ ~ bits2uint(bitsinvert(bytes2bits(bytes0)) +1) + 2 ^ (8 * n→i(# bytes0)) = bits2uint(bytes2bits(bytes0)), bytes2bits(bytes) + bytes2bits(bytes0) = [], bytes0 = [], bytes = [], bytes2bits(bytes).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 95:
0 < bits2uint(bitsinvert(bytes2bits(bytes0)) +1), bits2uint(bitsinvert(bytes2bits(bytes0)) +1) ≤ 2 ^ (8 * n→i(# bytes0)), bytes2bits(bytes0).first ⊦ 2 ^ (8 * n→i(# bytes0)) = bits2uint(bytes2bits(bytes0)) + bits2uint(bitsinvert(bytes2bits(bytes0)) +1), bytes = [], bytes0 = [], bytes2bits(bytes) + bytes2bits(bytes0) = [], bytes2bits(bytes).first

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ n→i(n);% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ ~ i < 0 ↔ 0 < i;% ⊦ i + 0 = i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;