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Proof step 140 from proof bxor-flip-one-bit (previous step)

0 ≤ j, int2bits(i) = (bits1 + boolvar ') + bits, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))), 2 ^ (j + 1) < i, 0 ≤ i, 0 ≤ j ⊦ bits = [], bits1 = [], bits1.first ↔ ¬ (bits1.rest + (¬ boolvar)').first, (sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1] ↔ ¬ boolvar) ∧ sgn((bits1 + (¬ boolvar)') + bitsxor(bits, mklist(false, i→n(j))))[# bits1, boolvar] = int2bits(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ 0 < i ∧ 0 ≤ j → (2 ^ j < i ↔ i→n(j) + 2 ≤ # int2bits(i) ∧ i ≠ 2 ^ j) subst: [i, j] → [i, j + 1] rotate: F spec: int2bits inst: name: int2bits-lslen-12

Additional information

Used simplifier rules: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ int2bits(0) = false ';%⊦ bits + boolvar ' + bits0 = boolvar0 ' ↔ bits = [] ∧ (boolvar ↔ boolvar0) ∧ bits0 = []; Used terms: i, j + 1