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Proof step 10 from proof bxor-flip-one-bit (previous step)

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