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Proof step 24 from proof and-first (previous step)

bitsand(boolvar ' + bits1, boolvar0 ' + bits) = boolvar1 ' + bits0, ∀ bits2, boolvar, bits1, boolvar0, boolvar1, bits0. # bits2 < # bits → bitsand(boolvar ' + bits1, boolvar0 ' + bits2) = boolvar1 ' + bits0 → (boolvar1 ↔ boolvar ∧ boolvar0) ⊦ boolvar1 ↔ boolvar ∧ boolvar0, boolvar0 ' + bits = true ', boolvar0 ' + bits = false ', boolvar ' + bits1 = true ', boolvar ' + bits1 = false ', bits1 ≠ [] ∧ bits ≠ []

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 6

Additional information

Used simplifier rules: ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)