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Proof step 3 from proof del-e (previous step)

⊦ (∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ a) ↔ (∀ a0. ¬ a0 ∈ s) ∧ a = b, a ∈ s

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a = b ⊦ (∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ a) ↔ (∀ a0. ¬ a0 ∈ s) ∧ a = b, a ∈ s
Proof step 10:
⊦ (∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ a) ↔ (∀ a0. ¬ a0 ∈ s) ∧ a = b, a = b, a ∈ s

Rule argument

fma: a = b

Additional information

Used formulas: a = b