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Proof step 48 from proof bytes2int-append (previous step)

bytes2bits(bytes).first, bits2int(bytes2bits(bytes0)) < 0 ⊦ bytes = [], bytes0 = [], bytes2bits(bytes) + bytes2bits(bytes0) = [], bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)) = [], (bits2int(bytes2bits(bytes0)) + 1) % 2 ^ (8 * n→i(# bytes0)) = bits2int(bytes2bits(bytes0)) + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 49:
⊦ abs(i) < j → i % j = i

Rule argument

seq: ⊦ abs(i) < j → i % j = i subst: [j, i] → [2 ^ (8 * n→i(# bytes0)), bits2int(bytes2bits(bytes0)) + 1] rotate: F spec: int-div inst: name: mod-zero-05

Additional information

Used simplifier rules: bytes2bits-nls: bytes ≠ [] ⊦ ¬ 2 ^ (8 * n→i(# bytes)) + bits2int(bytes2bits(bytes)) < 0;% ⊦ i < 0 → abs(i + 1) = abs(i) - 1;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ ~ i - j = ~(i + j);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i; Used terms: 2 ^ (8 * n→i(# bytes0)), bits2int(bytes2bits(bytes0)) + 1