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Proof step 158 from proof bxor-flip-one-bit (previous step)

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))), bits1.first ↔ ¬ (bits1.rest + (¬ boolvar)').first, 2 ^ j * 2 < i, 0 ≤ i, 0 ≤ j ⊦ bits = [], bits1 = [], ((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: ⊦ n < # bits → (bits + bits0)[n, boolvar] = bits[n, boolvar] + bits0 subst: [boolvar, n, bits0, bits] → [boolvar, # bits1, bitsxor(bits, mklist(false, i→n(j))), bits1 + (¬ boolvar)'] rotate: F spec: list-del inst: bitlista name: put-app-left

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: boolvar, # bits1, bitsxor(bits, mklist(false, i→n(j))), bits1 + (¬ boolvar)'