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Proof step 9 from proof bitsxor-zeros (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ # bits = 1 ↔ bits = bits.first + []
Proof step 11:
bits0 = bits0.first + [], 1 ≤ # bits, ∀ bits, bits1. # bits1 < 1 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ bits = [], bits0 = [], bitsxor(bits, bits0) = bits, true ∈ bits0

Rule argument

seq: ⊦ # bits = 1 ↔ bits = bits.first + [] subst: [bits] → [bits0] rotate: F spec: list-data inst: bitlista name: lf-01

Additional information

Used terms: bits0