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Proof step 5 from proof del-eq (
previous step
)
a2 ∈ s, a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a2 ∈ s ↔ a2 ∈ s0
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
a2 ∈ s, a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a2 ∈ s0
Rule argument
Additional information