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Proof step 20 from proof bitsand-app-false (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ # 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 ';% ⊦ (boolvar ' + bits + boolvar0 ').last ↔ boolvar0;% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;