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Proof step 7 from proof int2bytes-leading-zeros (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
int2bytes(i) = i2b(0) ' + i2b(0) ' + bytes, 0 ≤ i, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

Proof step 7 from proof int2bytes-leading-zeros (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information