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Proof step 65 from proof no-leading-zero (previous step)

int2bytes(bytes2int(i2b(0) ' + by ' + bytes1)) = i2b(0) ' + by ' + bytes1, int2bytes(band(b2i(by0), 255)).last = by0, band(b2i(by0), 255) < 256, 255 < bytes2int(i2b(0) ' + by ' + bytes1) * 256 + band(b2i(by0), 255), 0 ≤ b2i(by), 0 < bytes2int(i2b(0) ' + by ' + bytes1) * 256 + band(b2i(by0), 255), 0 ≤ bytes2int(i2b(0) ' + by ' + bytes1), 0 ≤ band(b2i(by0), 255), ∀ by, bytes0. # bytes0 < #(bytes1 + by0 ') → int2bytes(bytes2int(i2b(0) ' + by ' + bytes0)) = i2b(0) ' + by ' + bytes0 ∧ 0 ≤ b2i(by) → false ⊦

User applied rule apply induction

And closed the goal!

Rule argument

subst: [by, bytes0] → [by, bytes1]

Additional information

Used simplifier rules: ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ # by ' = 0 +1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: by, bytes1