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Proof step 28 from proof bitsand-first-samesize (previous step)

bits = bits3 + boolvar2 ', # bits2 + 1 = # bits, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < # bits → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ # bits = 1, bitsand(boolvar ' + bits2 + boolvar1 ', boolvar0 ' + bits) = (boolvar ∧ boolvar0)' + bitsand(bits2, bits3) + (boolvar1 ∧ boolvar2)', bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
# bits2 = # bits3, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < #(bits3 + boolvar2 ') → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bitsand(boolvar ' + bits2 + boolvar1 ', boolvar0 ' + bits3 + boolvar2 ') = (boolvar ∧ boolvar0)' + bitsand(bits2, bits3) + (boolvar1 ∧ boolvar2)', bits3 = []

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar ' ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;