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Proof step 2 from proof bitsor-bi-ones (previous step)

⊦ bits = [], bits0 = [], bits2int(bitsor(bits, bits0) + bits1) = bits2int(true ' + bits1), false ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ false ∈ bits ∧ ¬ bits0.first ∧ bits ≠ [] ∧ bits0 ≠ [] → bitsor(bits, bits0) = bits
Proof step 4:
¬(¬ false ∈ bits ∧ ¬ bits0.first ∧ bits ≠ [] ∧ bits0 ≠ []) ⊦ bits = [], bits0 = [], bits2int(bitsor(bits, bits0) + bits1) = bits2int(true ' + bits1), false ∈ bits
Proof step 28:
¬ false ∈ bits ∧ ¬ bits0.first ∧ bits ≠ [] ∧ bits0 ≠ [] ⊦ bits = [], bits0 = [], bits2int(bits + bits1) = bits2int(true ' + bits1), false ∈ bits

Rule argument

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

Additional information

Used terms: bits0, bits