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Proof step 4 from proof sub-one-01 (previous step)

a1 ∈ s, ∀ a0. a0 ∈ s → a0 = a ⊦ a0 ∈ s ↔ a0 = a

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 5:
a1 ∈ s, a0 ∈ s → a0 = a, ∀ a0. a0 ∈ s → a0 = a ⊦ a0 ∈ s ↔ a0 = a

Rule argument

pos: left 2 terms: a0

Additional information

Used terms: a0