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Proof step 29 from proof bxor-flip-one-bit (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsxor(bits, bits0) ≠ []

Rule argument

name: bitsxor-empty rotate: F subst: [bits0, bits] → [false ' + true ' + mklist(false, i→n(j)), int2bits(i)]

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: false ' + true ' + mklist(false, i→n(j)), int2bits(i)