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Proof step 13 from proof bitsor-bi-zeroes (previous step)

# bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int(bitsor(bits3 + bits0, false ' + bits2) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits3 = [], true ∈ bits3, true ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ # bits0 = # bits2 ∧ bits ≠ [] ∧ bits0 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits0, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits0, bits2)
Proof step 15:
¬(# bits0 = # bits2 ∧ bits3 ≠ [] ∧ bits0 ≠ [] ∧ false ' ≠ [] ∧ bits2 ≠ []), # bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int(bitsor(bits3 + bits0, false ' + bits2) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits3 = [], true ∈ bits3, true ∈ bits0
Proof step 30:
# bits0 = # bits2 ∧ bits3 ≠ [] ∧ bits0 ≠ [] ∧ false ' ≠ [] ∧ bits2 ≠ [], # bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int((bitsor(bits3, false ') + bitsor(bits0, bits2)) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits3 = [], true ∈ bits3, true ∈ bits0

Rule argument

name: bitsor-app rotate: F subst: [bits2, bits1, bits0, bits] → [bits2, false ', bits0, bits3]

Additional information

Used terms: bits2, false ', bits0, bits3