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Proof step 2 from proof sub-trans (previous step)

a ∈ s, ∀ a. a ∈ s0 → a ∈ s1, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s1

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 3:
a ∈ s, ∀ a. a ∈ s0 → a ∈ s1, a ∈ s → a ∈ s0, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s1

Rule argument

pos: left 3 terms: a

Additional information

Used terms: a