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Proof step 15 from proof bitsor-sgn-in (previous step)

true ∈ bits ⊦ bits0 = [], bits = [], boolvar ↔ boolvar ∈ bitsor(bits, bits0), false ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ ¬ false ∈ bits ∧ ¬ bits0.first ∧ bits ≠ [] ∧ bits0 ≠ [] → bitsor(bits, bits0) = bits
Proof step 17:
¬(¬ false ∈ bits ∧ ¬ bits0.first ∧ bits ≠ [] ∧ bits0 ≠ []), true ∈ bits ⊦ bits0 = [], bits = [], boolvar ↔ boolvar ∈ bitsor(bits, bits0), false ∈ bits

Rule argument

name: bitsor-ones-false rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

Used terms: bits0, bits