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Proof step 34 from proof bxor-flip-one-bit (previous step)

0 ≤ i ∧ 1 < i→n(j), 2 ^ (n→i(i→n(j)) - 2) ≤ i ∧ i < 2 ^ (n→i(i→n(j)) - 1), 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))) = [], bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))).first ∧ sgn(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))))[0, false] = int2bits(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ i;% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i - j < i0 ↔ i < j + i0;%0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;% ⊦ 0 < j → (j < 2 ↔ j = 1);% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 0 ≤ i → n→i(i→n(i)) = i;%0 ≤ i ⊦ i→n(i) = (* m) ↔ i = n→i(m);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);