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Proof step 6 from proof int2bytes-first-zerobyte-01 (previous step)

n→i(# int2bits(j)) = i0 * 8 + j0, j0 < 8, 0 ≤ j, 0 ≤ j0 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
0 ≤ i ⊦ n→i(n) = (* i) ↔ n = i→n(i)
Proof step 8:
¬ 0 ≤ i0 * 8 + j0, n→i(# int2bits(j)) = i0 * 8 + j0, j0 < 8, 0 ≤ j, 0 ≤ j0 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)
Proof step 11:
0 ≤ i0 * 8 + j0, # int2bits(j) = i→n(i0 * 8 + j0), j0 < 8, 0 ≤ j, 0 ≤ j0 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

Rule argument

seq: 0 ≤ i ⊦ n→i(n) = (* i) ↔ n = i→n(i) subst: [i, n] → [i0 * 8 + j0, # int2bits(j)] rotate: F spec: intnat inst: name: num-n2i-eq

Additional information

Used terms: i0 * 8 + j0, # int2bits(j)