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Proof step 34 from proof uint2bits-add-high (previous step)

i→n(i) - 1 = 0, 0 < i, 2 ^ (i - 1) + 2 < 2 ^ i + 2 ^ i, 2 < 2 ^ i + 2 ^ i, 2 ^ (i - 1) + 1 < 2 ^ i, 1 < 2 ^ i, 2 ^ (i - 1) < 2 ^ i, 2 ^ (i - 1) + 2 < 2 ^ i + i, 2 < 2 ^ i + i, 0 ≤ 2 ^ i + 2 ^ i, 0 < 2 ^ i, 0 ≤ 2 ^ i, 0 ≤ 2 ^ i + i, ∀ i0, j. i→n(i0) < i→n(i) → true → j < 0 ∧ 0 < i0 ∧ 2 ^ (i0 - 1) < abs(j) ∧ abs(j) ≤ 2 ^ i0 → uint2bits(j + 2 ^ i0) = sgn(int2bits(j).rest) ⊦ i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ 0 < i → i→n(i) = i→n(i - 1) + 1 subst: [i] → [i] rotate: F spec: intnat inst: name: Int2nat-pos

Additional information

Used simplifier rules: ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ 0 + m = m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i = j → i - j = 0;% ⊦ i + j = j + i;%i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i + j + i0 = (i + j) + i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ j = i0 → i < i0; (forward) Used terms: i