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Proof step 9 from proof del-eq (
previous step
)
∀ a. a ∈ s ↔ a ∈ s0, a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s ⊦ ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 10
:
a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦ a2 ∈ s ↔ a2 ∈ s0, a2 = a
Rule argument
Additional information