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Proof step 28 from proof bitsand-onein (previous step)

1 < #(bits + false ' + bits3 + boolvar ') ∧ 1 < #(bits4 + boolvar0 '), ∀ bits0, bits, bits2. # bits2 < #(bits4 + boolvar0 ') → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ true ∈ bitsand((bits + false ' + bits3 + boolvar ').butlast, (bits4 + boolvar0 ').butlast) + ((bits + false ' + bits3 + boolvar ').last ∧ (bits4 + boolvar0 ').last) ↔ true ∈ bitsand(false ' + bits3, bits4) ∨ boolvar ∧ boolvar0, bits4 = [], true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
¬ boolvar ∨ ¬ boolvar0, ∀ bits0, bits, bits2. # bits2 < #(bits4 + boolvar0 ') → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ bits4 = [], true ∈ bitsand(bits + false ' + bits3, bits4) ↔ true ∈ bitsand(false ' + bits3, bits4), true ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (boolvar ' + bits + boolvar0 ').last ↔ boolvar0;% ⊦ (bits + boolvar ' + bits0).last ↔ (boolvar ' + bits0).last;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;%⊦ (bits + boolvar ' + bits0).butlast = bits + (boolvar ' + bits0).butlast;