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Proof step 121 from proof bxor-flip-one-bit (previous step)

0 ≤ j, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (false ' + true ') + bitsxor(bits, mklist(false, i→n(j))), int2bits(i) = false ' + bits, 0 ≤ j, false ∈ int2bits(i), 2 ^ (j + 1) < i, 0 < i ⊦ bits = [], # int2bits(i) = 2 + i→n(j), i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 122:
⊦ 0 < i ∧ 0 ≤ j → (2 ^ j < i ↔ i→n(j) + 2 ≤ # int2bits(i) ∧ i ≠ 2 ^ j)
Proof step 123:
0 < i ∧ 0 ≤ j + 1, 0 ≤ j, # bits = i→n(j), bitsxor(int2bits(i), (false ' + true ') + mklist(false, i→n(j))) = (false ' + true ') + bitsxor(bits, mklist(false, i→n(j))), int2bits(i) = false ' + bits, 0 ≤ j, false ∈ int2bits(i), i→n(j + 1) + 2 ≤ # int2bits(i) ∧ i ≠ 2 ^ (j + 1), 0 < i ⊦ bits = [], # int2bits(i) = 2 + i→n(j), i = 0

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;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < j → ¬ j < i;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; Used terms: i, j + 1