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Proof step 23 from proof and-first (previous step)

¬(boolvar0 ' + bits ≠ [] ∧ boolvar ' + bits1 = false '), 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 ', bits1 ≠ [] ∧ bits ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
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 ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];