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Proof step 2 from proof and-insgn (previous step)

⊦ bits = [], bits0 = [], boolvar ∈ bitsand(sgn(bits), bits0) ↔ boolvar ∈ bitsand(bits, bits0)

User applied rule induction

And obtained the following new goals

Proof step 3:
true, ∀ bits1, bits, boolvar. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ bits = [] ∨ bits0 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits0) ↔ boolvar ∈ bitsand(bits, bits0))

Rule argument

indvar: # bits0 wfpred: <

Additional information

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