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Proof step 27 from proof int2bits-len-range (previous step)

(2 ^ i0 + j) + 1 < i, 0 < 2 ^ i0, 0 ≤ 2 ^ i0, j < 0, 2 ^ i0 + j < 0, 0 < 2 ^ i0 * 2 + j, j < 2 ^ i0, j < i, j < 2 ^ i, 2 ^ i0 + j ≤ i, 0 ≤ i, 2 ^ i0 + j < 2 ^ i, j + 1 < 2 ^ i0, 0 < 2 ^ i, j < i0, i0 ≤ i, 2 ^ i0 + j ≤ i0, i→n(i) + 1 < i→n(i0 + 2), 0 ≤ 2 ^ i + j, 0 ≤ i0, 2 ^ i0 + j < i, i0 < i, 2 ^ i0 + j < i0, 1 < 2 ^ i0, (2 ^ i0 + j) + 1 < 2 ^ i, j < ~1, 0 < i ⊦ 2 ^ i0 + j = 0, j = 0, ~2 = i0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j) subst: [j, i] → [2, i0] rotate: F spec: intnat inst: name: add-01

Additional information

Used simplifier rules: ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;%n < n0 ⊦ m + (* n) < m0 + (* n0) ↔ m < m0 + (n0 - n);% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);% ⊦ 0 ≤ i → abs(i) = i;% ⊦ 0 < j → abs(j) = j;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i < j → ¬ j < i; Used terms: 2, i0