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Proof step 12 from proof bxor-flip-one-bit (previous step)

j = i0 + 2, 0 < j, 2 < j, 2 ≤ j, 0 ≤ i, 2 ^ j * 2 < i, i < 2 ^ i0, 0 < i ⊦ j = 0, ( int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j)))))[# int2bits(i) - i→n(j) + 1] ↔ ¬ int2bits(i)[# int2bits(i) - i→n(j) + 1]) ∧ int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))))) [# int2bits(i) - i→n(j) + 1, int2bits(i)[# int2bits(i) - i→n(j) + 1] ] = int2bits(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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