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Proof step 11 from proof uint2bits-add-low (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: j = 1 - 2 * 2 ^ (i - 1)

Additional information

Used simplifier rules: int2bits-neg-num: i < 0 ⊦ int2bits((* i)) = sgn(bitsinvert(uint2bits(abs(i))) +1);%uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i * j = j * i;% ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i - j ≤ i0 ↔ i ≤ j + i0;% ⊦ i + 0 = i;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ (* i0) * i + i = (i0 + 1) * i;% ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ (* i2) * i ≤ i + j0 ↔ (i2 - 1) * i ≤ j0;% ⊦ 1 * i = i;% ⊦ 2 ^ i ≤ 1 ↔ i < 1;% ⊦ 0 < j → (j < 2 ↔ j = 1);% ⊦ i = j → i - j = 0;% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ bitsinvert(false ' + bits) = true + bitsinvert(bits);% ⊦ bitsinvert(true ') = false ';% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + false ')+1 = bits + true;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ sgn(true ' + true ') = true ';% ⊦ 0 < i → fillsgn(i→n(i), boolvar ') = mklist(boolvar, i→n(i));% ⊦ mklist(boolvar, n) = boolvar ' ↔ n = 1;%0 ≤ i ⊦ i→n(i) = (* m) ↔ i = n→i(m); Used formulas: j = 1 - 2 * 2 ^ (i - 1)