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Proof step 7 from proof union-write (previous step)

n < # asar, disj(asar, 0, n + 1), ∀ n0, asar, as. n0 < n ∧ n0 < # asar ∧ disj(asar, 0, n) ∧ n ≤ # asar → union(asar[n0, as], 0, n) = (union(asar, 0, n) \ asar[n0]) ∪ as ⊦ as ∪ union(asar, 0, n) = ((asar[n] ∪ union(asar, 0, n)) \ asar[n]) ∪ as

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ (as ∪ as0) \ as1 = (as \ as1) ∪ (as0 \ as1)
Proof step 9:
n < # asar, disj(asar, 0, n + 1), ∀ n0, asar, as. n0 < n ∧ n0 < # asar ∧ disj(asar, 0, n) ∧ n ≤ # asar → union(asar[n0, as], 0, n) = (union(asar, 0, n) \ asar[n0]) ∪ as ⊦ as ∪ union(asar, 0, n) = ((asar[n] \ asar[n]) ∪ (union(asar, 0, n) \ asar[n])) ∪ as

Rule argument

spec: set-union, inst: address-list-set name: diff-union-03 seq: ⊦ (as ∪ as0) \ as1 = (as \ as1) ∪ (as0 \ as1) subst: [as1, as0, as] → [asar[n], union(asar, 0, n), asar[n]] flags: <all paths>

Additional information

Used terms: asar[n], union(asar, 0, n), asar[n]