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Proof step 16 from proof bitsxor-is-orand (previous step)

∀ bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ # bits0 = 1, bits0 = false ', bits0 = true ', bits0 = [], bits2int(bitsxor(true ', bits0)) = bits2int(bitsor(false ', bitsand(bitsinvert(bits0), true ')))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ bits ≠ [] ∧ bits0 = true ' → bitsand(bits, bits0) = bits
Proof step 18:
bitsinvert(bits0) ≠ [] ∧ true ' = true ', ∀ bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ # bits0 = 1, bits0 = false ', bits0 = true ', bits0 = [], bits2int(bitsxor(true ', bits0)) = bits2int(bitsor(false ', bitsinvert(bits0)))

Rule argument

name: and-rec-01 rotate: F subst: [bits0, bits] → [true ', bitsinvert(bits0)]

Additional information

Used simplifier rules: ⊦ bitsinvert(bits) = [] ↔ bits = []; Used terms: true ', bitsinvert(bits0)