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Proof step 28 from proof bitsor-first-02 (previous step)

bitsor(false ' + bits3, false ' + bits0) = [], boolvar ∨ boolvar0, ∀ bits1, bits2, bits. # bits < #(bits3 + boolvar ') → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ ¬ boolvar ∧ ¬ boolvar0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsor(bits, bits0) ≠ []

Rule argument

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

Additional information

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