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Proof step 12 from proof bitsand-onein (previous step)

¬(false ' + bits0 ≠ [] ∧ bits1 = false '), ∀ bits0, bits, bits2. # bits2 < # bits1 → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ true ∈ bitsand(bits + false ' + bits0, bits1) ↔ true ∈ bitsand(false ' + bits0, bits1), bits1 = [], bits1 = true ', true ∈ bits, bits0 ≠ [] ∧ # bits1 ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
true ∈ bitsand(bits + false ', bits1), ∀ bits0, bits, bits2. # bits2 < # bits1 → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ # bits1 = 1, bits1 = false ', bits1 = true ', bits1 = [], true ∈ bits

Rule argument

Additional information

Used simplifier rules: and-zero-02: ⊦ bits ≠ [] → bitsand(false ', bits) = false ';% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ bits + [] = bits;% ⊦ # bits = 1 ∧ bits ≠ true ' → bits = false '; (forward)% ⊦ # bits = 1 ∧ bits ≠ false ' → bits = true '; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];