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Proof step 26 from proof bitsor-sgn-in (previous step)

# bits0 = # bits1, true ∈ bits2 ⊦ bits2 + bits0 = [], boolvar ↔ boolvar ∈ bitsor(bits2 + bits0, true ' + bits1), # bits2 = 1, bits2 = [], false ∈ bits2, false ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ # bits0 = # bits2 ∧ bits ≠ [] ∧ bits0 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits0, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits0, bits2)
Proof step 28:
¬(# bits0 = # bits1 ∧ bits2 ≠ [] ∧ bits0 ≠ [] ∧ true ' ≠ [] ∧ bits1 ≠ []), # bits0 = # bits1, true ∈ bits2 ⊦ bits2 + bits0 = [], boolvar ↔ boolvar ∈ bitsor(bits2 + bits0, true ' + bits1), # bits2 = 1, bits2 = [], false ∈ bits2, false ∈ bits0
Proof step 33:
# bits0 = # bits1 ∧ bits2 ≠ [] ∧ bits0 ≠ [] ∧ true ' ≠ [] ∧ bits1 ≠ [], # bits0 = # bits1, true ∈ bits2 ⊦ bits2 + bits0 = [], boolvar ↔ boolvar ∈ bitsor(bits2, true ') + bitsor(bits0, bits1), # bits2 = 1, bits2 = [], false ∈ bits2, false ∈ bits0

Rule argument

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

Additional information

Used terms: bits1, true ', bits0, bits2