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Proof step 40 from proof bitsxor-is-orand (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 41:
⊦ bits ≠ [] ∧ bits0 = true ' → bitsxor(bits, bits0) = bitsinvert(bits)
Proof step 42:
bits ≠ [] ∧ true ' = true ', ∀ bits, bits1. # bits1 < # true ' → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bits = [], bits2int(bitsinvert(bits)) = bits2int(bitsor(bitsinvert(bits), false '))

Rule argument

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

Additional information

Used terms: true ', bits