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Proof step 6 from proof no-leading-zero-01 (previous step)

¬ 255 < i, int2bytes(i) = i2b(0) ' + by ' + bytes, i2b(0) ∈ int2bytes(i), 0 ≤ b2i(by), ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + by ' + bytes ∧ 0 ≤ b2i(by) ∧ i2b(0) ∈ int2bytes(i0) → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
int2bytes(i) = i2b(0) ' + by ' + bytes, i2b(0) ∈ int2bytes(i), 0 ≤ b2i(by), i ≤ 255, ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + by ' + bytes ∧ 0 ≤ b2i(by) ∧ i2b(0) ∈ int2bytes(i0) → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;