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Proof step 8 from proof bitsor-first-01 (previous step)

bitsor(boolvar ' + bits1, boolvar0 ' + bits) = [] ⊦ [].first ↔ boolvar ∨ boolvar0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsor(bits, bits0) ≠ []

Rule argument

name: bitsor-empty rotate: F subst: [bits0, bits] → [boolvar0 ' + bits, boolvar ' + bits1]

Additional information

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