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Proof step 4 from proof bitsxor-zeros (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsxor(bits, bits0) = bitsxor(bits.butlast, bits0.butlast) + (¬(bits.last ↔ bits0.last))
Proof step 6:
¬(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
Proof step 16:
1 < # bits ∧ 1 < # bits0, # bits0 ≤ # bits, ∀ bits, bits1. # bits1 < # bits0 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ bitsxor(bits.butlast, bits0.butlast) + (¬(bits.last ↔ bits0.last)) = bits, bits0 = [], bits = [], true ∈ bits0

Rule argument

name: bitsxor-rec-01 rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

Used terms: bits0, bits