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Proof step 6 from proof add-ones-grlen-02 (previous step)

¬(1 < # bits ∧ 1 < # bits0), # bits0 ≤ # bits, ∀ bits, bits1. # bits1 < # bits0 → # bits1 ≤ # bits → bits1 = [] ∨ bitsand(bits, bits1) = bits1 ∨ false ∈ bits ∨ bits1.first ⊦ bitsand(bits, bits0) = bits0, bits0 = [], bits0.first, false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bits0 ≤ # bits, ∀ bits, bits1. # bits1 < # bits0 → # bits1 ≤ # bits → bits1 = [] ∨ bitsand(bits, bits1) = bits1 ∨ false ∈ bits ∨ bits1.first ⊦ bitsand(bits, bits0) = bits0, bits0 = [], bits = [], bits0.first, false ∈ bits, # bits0 ≠ 1 ∧ # bits ≠ 1

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ 0 ↔ m = 0;% ⊦ # [] = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);