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Proof step 6 from proof selsub-diff (
previous step
)
∀ a. a ∈ s1 → a ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2, (a0 ∈ s2 ∧ ¬ a0 ∈ s1 → a0 ∈ s2) ∧ (a1 ∈ s1 ↔ a1 ∈ s2 \ s2 \ s1)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 7
:
∀ a. a ∈ s1 → a ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2, a1 ∈ s1 ↔ a1 ∈ s2 \ s2 \ s1
Rule argument
Additional information