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Proof step 2 from proof sub-add-03 (previous step)

∀ a0. a0 = a ∨ a0 ∈ s → a0 ∈ s0 ⊦ (a0 ∈ s → a0 ∈ s0) ∧ a ∈ s0

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 3:
a = a ∨ a ∈ s → a ∈ s0, ∀ a0. a0 ∈ s → a0 ∈ s0 ⊦ (a0 ∈ s → a0 ∈ s0) ∧ a ∈ s0

Rule argument

pos: left 1 terms: a

Additional information

Used terms: a