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Proof step 16 from proof bitsand-onein (previous step)

true ∈ bitsand(bits + false ' + bits0, false '), ∀ bits0, bits, bits2. # bits2 < # false ' → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ true ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ bits ≠ [] ∧ bits0 = false ' → bitsand(bits, bits0) = false '

Rule argument

name: and-rec-02 rotate: F subst: [bits0, bits] → [false ', bits + false ' + bits0]

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ bits + boolvar ' + bits0 ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0; Used terms: false ', bits + false ' + bits0