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Proof step 128 from proof bxor-flip-one-bit (previous step)

int2bits(i) = false ' + bits0, # int2bits(i) = 2 + i→n(j), # bits0 = i→n(j), bitsxor(bits0, true ' + mklist(false, i→n(j))) = true ' + bitsxor(bits0, mklist(false, i→n(j))), 0 ≤ j, 2 ^ j * 2 < i, 0 < i, false ∈ int2bits(i) ⊦ bits0 = [], i = 0

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 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;% ⊦ m = m + n ↔ n = 0;