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Proof step 14 from proof and-size-one-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# bits3 = # bits2, ∀ bits, bits1, bits0. # bits < #(bits3 + boolvar ') → # bits = # bits1 → bits1 = [] ∨ bits0 = [] ∨ bitsand(true ' + bits, bits0 + bits1) = bits0 + bitsand(bits, bits1) ⊦ bits0 = [], bitsand(true ' + bits3, bits0 + bits2) + (boolvar ∧ boolvar0)' = bits0 + bitsand(bits3 + boolvar ', bits2 + boolvar0 '), bits3 ≠ [] ∧ bits2 ≠ []

Rule argument

Additional information

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