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Proof step 17 from proof bitsand-app-ones (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# bits < # bits1, ∀ bits0, bits, bits2. # bits2 < # bits1 → true → bitsand(bits + true ' + bits0, false ' + bits2) = bitsand(true ' + bits0, false ' + bits2) ∨ false ∈ bits ⊦ bitsand(bits + true ', false ' + bits1) = false ' + bits1, bits1 = [], false ∈ bits

Rule argument

Additional information

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