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Proof step 1 from proof and-true (conclusion)

⊦ bits ≠ [] ∧ bits0 ≠ [] → (bitsand(bits, bits0) = true ' ↔ bits = true ' ∧ bits0 = true ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bitsand(bits, bits0) = true ' ⊦ bits0 = [], bits = [], bits = bitsand(bits, bits0) ∧ bits0 = bitsand(bits, bits0)

Rule argument

Additional information

Used simplifier rules: and-onel-01: ⊦ bits ≠ [] → bitsand(true ', bits) = bits;%and-oner-01: ⊦ bits ≠ [] → bitsand(bits, true ') = bits;% ⊦ boolvar ' ≠ [];% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;