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Proof step 23 from proof int2bytes-first-zerobyte-01 (previous step)

1 = i→n(j0) + 8 * i→n(i0), 0 ≤ i0, 0 ≤ j0, j0 < 8 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ 0 < i)

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 24:
⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1)
Proof step 25:
1 < 8 ∧ i→n(j0) < 8 → (0 * 8 + 1 = i→n(i0) * 8 + i→n(j0) ↔ 0 = i→n(i0) ∧ 1 = i→n(j0)), 1 = i→n(j0) + 8 * i→n(i0), 0 ≤ i0, 0 ≤ j0, j0 < 8 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ 0 < i)

Rule argument

seq: ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1) subst: [n0, n, n1, m, m0] → [1, 8, i→n(j0), 0, i→n(i0)] rotate: F spec: nat-mult inst: name: modeq

Additional information

Used terms: 1, 8, i→n(j0), 0, i→n(i0)