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Proof step 8 from proof subseteq-union-diff-one (previous step)

n0 < # asar, n < # asar, disj(asar, 0, # asar), adr ∈ asar[n], n ≠ n0 ∨ (∀ adr. adr ∈ asar[n0] → adr ∈ as) ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ as, adr ∈ union(asar, 0, # asar) \ asar[n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ adr ∈ as \ as0 ↔ adr ∈ as ∧ ¬ adr ∈ as0
Proof step 10:
n < # asar, n0 < # asar, disj(asar, 0, # asar), adr ∈ asar[n], n ≠ n0 ∨ (∀ adr. adr ∈ asar[n0] → adr ∈ as) ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ as, adr ∈ union(asar, 0, # asar) ∧ ¬ adr ∈ asar[n0]

Rule argument

spec: set-union, inst: address-list-set name: Difference seq: ⊦ adr ∈ as \ as0 ↔ adr ∈ as ∧ ¬ adr ∈ as0 subst: [as0, as, adr] → [asar[n0], union(asar, 0, # asar), adr] flags: <all paths>

Additional information

Used terms: asar[n0], union(asar, 0, # asar), adr