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Proof step 1 from proof bitsor-bi-sgn (conclusion)

⊦ bits ≠ [] ∧ bits0 ≠ [] → bits2int(bitsor(sgn(bits), bits0)) = bits2int(bitsor(bits, bits0))

User applied rule induction

And obtained the following new goals

Proof step 2:
true, ∀ bits, bits1. # bits1 < # bits0 → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits ≠ [] ∧ bits0 ≠ [] → bits2int(bitsor(sgn(bits), bits0)) = bits2int(bitsor(bits, bits0))

Rule argument

indvar: # bits0 wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: # bits0