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Proof step 14 from proof inter-union-diff-subseteq (previous step)

adr ∈ as3, as1 ∩ as3 ⊆ as, ∀ adr. adr ∈ as0 ∩ as3 → adr ∈ as, adr ∈ as0 ∨ adr ∈ as1 ∧ ¬ adr ∈ as2 ⊦ # as3 = 0, #(as0 ∪ (as1 \ as2)) = 0, as3 = ∅, as0 ∪ (as1 \ as2) = ∅, adr ∈ as

Heuristic cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: adr ∈ as0

Additional information

Used formulas: adr ∈ as0