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Proof step 10 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) = (union(asar, 0, n) \ asar[n]) ∪ as

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ as ∩ as0 = ∅ → as \ as0 = as
Proof step 12:
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 ⊦ union(asar, 0, n) ∩ asar[n] = ∅, as ∪ union(asar, 0, n) = (union(asar, 0, n) \ asar[n]) ∪ as

Rule argument

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

Additional information

Used simplifier rules: unc: ⊦ as ∪ as0 = as0 ∪ as; from specification set-union[address-list-set] Used terms: asar[n], union(asar, 0, n)