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Proof step 43 from proof int2bits-lslen-07 (previous step)

0 ≤ i + 1, 0 ≤ i, 0 < j, 0 ≤ i, 2 ^ i ≤ j, i→n(i + 2) + 1 < #(false ' + true ' + mklist(false, i→n(i + 1))) ⊦ 0 = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
i→n(i + 2) < i→n(i) + 2, 2 ^ i ≤ j, 0 ≤ i, 0 < j ⊦ 0 = j

Rule argument

Additional information

Used simplifier rules: n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ m < n → n ≠ 0; (forward)%n < n0 ⊦ m + (* n) < m0 + (* n0) ↔ m < m0 + (n0 - n);% ⊦ # mklist(boolvar, n) = n;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ mklist(boolvar, 1) = boolvar ';% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;