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Proof step 3 from proof in-union-ext (previous step)

∀ adr, asar, n. adr ∈ union(asar, n, m0) ↔ (∃ n0. n ≤ n0 ∧ n0 < n + m0 ∧ adr ∈ asar[n0]) ⊦ adr ∈ asar[n + m0] ∪ union(asar, n, m0) ↔ (∃ n0. n ≤ n0 ∧ ¬ n + m0 < n0 ∧ adr ∈ asar[n0])

User applied rule all left

And obtained the following new goals

Proof step 4:
adr ∈ union(asar, n, m0) ↔ (∃ n0. n ≤ n0 ∧ n0 < n + m0 ∧ adr ∈ asar[n0]) ⊦ adr ∈ asar[n + m0] ∪ union(asar, n, m0) ↔ (∃ n0. n ≤ n0 ∧ ¬ n + m0 < n0 ∧ adr ∈ asar[n0])

Rule argument

pos: left 1 terms: adr, asar, n abort: F discard: T computed: F

Additional information

Used terms: adr, asar, n