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Proof step 2 from proof del-e (
previous step
)
⊦ (∀ a0. a0 = b ∨ a0 ≠ a ∧ ¬ a0 ∈ s) ↔ (∀ a0. ¬ a0 ∈ s) ∧ a = b, a ∈ s
Heuristic cut applied rule cut formula
And obtained the following new goals
Proof step 3
:
⊦ a = b, (∀ a0. a0 = b ∨ a0 ≠ a ∧ ¬ a0 ∈ s) ↔ (∀ a0. ¬ a0 ∈ s) ∧ a = b, a ∈ s
Rule argument
fma: a = b
Additional information
Used formulas: a = b