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Proof step 124 from proof bxor-flip-one-bit (previous step)

# 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 < i, 0 ≤ j, false ∈ int2bits(i), 3 + i→n(j) ≤ # int2bits(i) ⊦ bits = [], # int2bits(i) = 2 + i→n(j), i = 0, i = 2 ^ j * 2

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 3 flags: discard

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n);% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;%n0 < n ⊦ m + (* n) ≤ m0 + (* n0) ↔ m + (n - n0) ≤ m0;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ m = m + n ↔ n = 0;% ⊦ m + m0 ≤ m ↔ m0 = 0;