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Proof step 21 from proof bitsor-first (previous step)

bitsor(true ' + bits3, bits4) = [], ∀ bits1, bits, bits2. # bits2 < #(bits4 + false ') → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ bits4 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsor(bits, bits0) ≠ []

Rule argument

name: bitsor-empty rotate: F subst: [bits0, bits] → [bits4, true ' + bits3]

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: bits4, true ' + bits3