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Proof step 14 from proof bitsand-app (previous step)

# bits1 = # bits4, ∀ bits2, bits0, bits, bits3. # bits3 < #(bits4 + boolvar ') → # bits = # bits3 → bits0 = [] ∨ bits2 = [] ∨ bits = [] ∨ bitsand(bits0 + bits, bits2 + bits3) = bitsand(bits0, bits2) + bitsand(bits, bits3) ⊦ bits2 = [], bits0 = [], bitsand(bits0 + bits1, bits2 + bits4) + (boolvar0 ∧ boolvar)' = bitsand(bits0, bits2) + bitsand(bits1 + boolvar0 ', bits4 + boolvar '), bits1 ≠ [] ∧ bits4 ≠ []

User applied rule cut formula

And closed the goal!

Rule argument

fma: bits1 ≠ []

Additional information

Used simplifier rules: and-one: ⊦ bitsand(boolvar ', boolvar0 ') = (boolvar ∧ boolvar0)';% ⊦ # [] = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1; Used formulas: bits1 ≠ []