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Proof step 16 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 ').butlast, (bits2 + boolvar0 ').butlast) + ((bits3 + boolvar ').last ∧ (bits2 + boolvar0 ').last))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# 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) = bits0 + bitsand(bits3, bits2), bits3 = [], bits2 = []

Rule argument

Additional information

Used simplifier rules: ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ # 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 + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + bits0 = bits1 + bits0 ↔ bits = bits1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;