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Proof step 4 from proof eq-01 (previous step)

∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a0 ∈ s ↔ a0 ∈ s0, a ∈ s0, a ∈ s

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 5:
a ≠ a → (a ∈ s ↔ a ∈ s0), ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a0 ∈ s ↔ a0 ∈ s0, a ∈ s0, a ∈ s

Rule argument

pos: left 1 terms: a

Additional information

Used terms: a