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Proof step 2 from proof unsubdel (previous step)

∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, a0 ∈ s ∨ a0 ∈ s1 ⊦ a ∈ s1, a ∈ s, a0 ≠ a ∧ (a0 ∈ s2 ∨ a0 ∈ s0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 3:
a0 ∈ s, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, a0 ∈ s ∨ a0 ∈ s1 ⊦ a ∈ s1, a ∈ s, a0 ≠ a ∧ (a0 ∈ s2 ∨ a0 ∈ s0)
Proof step 5:
∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, a0 ∈ s ∨ a0 ∈ s1 ⊦ a0 ∈ s, a ∈ s1, a ∈ s, a0 ≠ a ∧ (a0 ∈ s2 ∨ a0 ∈ s0)

Rule argument

fma: a0 ∈ s

Additional information

Used formulas: a0 ∈ s