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Proof step 145 from proof bxor-flip-one-bit (previous step)

bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (false ' + bits0) + (¬ boolvar)' + bitsxor(bits, mklist(false, i→n(j))), # bits = i→n(j), int2bits(i) = (false ' + bits0) + boolvar ' + bits, 0 < i, 3 + i→n(j) ≤ # int2bits(i), 0 ≤ j, 0 ≤ i, false ∈ bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))), false ∈ int2bits(i) ⊦ bits = [], i = 2 ^ j * 2, (bits0 + (¬ boolvar)').first, (sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1 + # bits0] ↔ ¬ boolvar) ∧ sgn(bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))))[1 + # bits0, boolvar] = int2bits(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: bitsxor-first rotate: F subst: [bits0, bits] → [(false ' + true ') + mklist(false, i→n(j)), int2bits(i)]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ # mklist(boolvar, n) = n;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ int2bits(j).first ↔ j < 0;% ⊦ i < j → ¬ j < i;% ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # bits = 0 ↔ bits = [];% ⊦ int2bits(j) ≠ [];% ⊦ 0 ≤ i → (# int2bits(i) = 1 ↔ i = 0);% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ m + n = n + m;%n0 < n ⊦ m + (* n) ≤ m0 + (* n0) ↔ m + (n - n0) ≤ m0;% ⊦ m + m0 ≤ m ↔ m0 = 0;% ⊦ (m + n0) + n1 = m + n0 + n1; Used terms: (false ' + true ') + mklist(false, i→n(j)), int2bits(i)