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Proof step 28 from proof int2bits-len-neg (previous step)

n→i(# bits) = i + 1, 0 < 2 ^ i * 2 + i, ~1 ≤ 2 ^ i * 2 + i, 2 ^ i < 2 * (2 ^ i + 2 ^ i), 2 ^ i + 1 < 2 * (2 ^ i + 2 ^ i), 0 < 2 ^ i + i, 0 < 2 ^ i, 0 ≤ 2 ^ i, 0 ≤ 2 ^ i + 2 ^ i, 1 < 2 ^ i * 2 + i, 0 ≤ i ⊦ # bits = 1 + i→n(i), bits = [], true ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
0 ≤ i ⊦ n→i(n) = i ↔ n = i→n(i)

Rule argument

seq: 0 ≤ i ⊦ n→i(n) = (* i) ↔ n = i→n(i) subst: [i, n] → [i + 1, # bits] rotate: F spec: intnat inst: name: num-n2i-eq

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) Used terms: i + 1, # bits