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Proof step 8 from proof bxor-flip-one-bit (previous step)

0 ≤ i, 2 ^ j * 2 < i, # int2bits(i) < i→n(j), 0 < j ⊦ j = 0, ( int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j)))))[# int2bits(i) - i→n(j) + 1] ↔ ¬ int2bits(i)[# int2bits(i) - i→n(j) + 1]) ∧ int2bits(bits2int(bitsxor(int2bits(i), false ' + true ' + mklist(false, i→n(j))))) [# int2bits(i) - i→n(j) + 1, int2bits(i)[# int2bits(i) - i→n(j) + 1] ] = int2bits(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ 0 < i → (# int2bits(i) < m ↔ i < 2 ^ (n→i(m) - 2)) subst: [m, i] → [i→n(j), i] rotate: F spec: int2bits inst: name: int2bits-lslen

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m + n = n + m;% ⊦ int2bits(0) = false ';% ⊦ # boolvar ' = 0 +1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;%0 ≤ i ⊦ i→n(i) = (* m) ↔ i = n→i(m);% ⊦ 0 < i → (i * j < 0 ↔ j < 0);% ⊦ ¬ 2 ^ i < 0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i; Used terms: i→n(j), i