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Proof step 105 from proof bxor-flip-one-bit (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 106:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsxor(bits, bits0) ≠ []

Rule argument

name: bitsxor-empty rotate: F subst: [bits0, bits] → [false ' + true ', bits0]

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: false ' + true ', bits0