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Proof step 5 from proof and-notone-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
bitsand(bits, false ' + bits0) = true ' + bits1, bits = bits.first ', ∀ bits2, bits0, bits1. # bits2 < # bits → bitsand(bits2, false ' + bits0) = true ' + bits1 → bits2 = [] ⊦ bits = [], bits0 = []

Rule argument

Additional information

Used simplifier rules: and-zero-03: ⊦ bits ≠ [] → bitsand(bits, false ') = false ';% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);