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Proof step 6 from proof and-ass (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ bits2, bits0, bits1. # bits2 < # bits → true → bits2 = [] ∨ bits0 = [] ∨ bits1 = [] ∨ bitsand(bits2, bitsand(bits0, bits1)) = bitsand(bitsand(bits2, bits0), bits1) ⊦ bitsand(bits, bitsand(bits0, bits1)) = bitsand(bitsand(bits, bits0), bits1), bits1 = [], bits0 = [], bits = [], bits0 ≠ bits0.first ' ∧ bits1 ≠ bits1.first '

Rule argument

Additional information

Used simplifier rules: ⊦ (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);