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Proof step 6 from proof bitsxor-zeros (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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