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Proof step 2 from proof sub-del-04 (
previous step
)
⊦ (∀ a0. a0 ∈ s → a0 ≠ a ∧ a0 ∈ s0) ↔ (∀ a. a ∈ s → a ∈ s0) ∧ ¬ a ∈ s
Heuristic cut applied rule cut formula
And obtained the following new goals
Proof step 3
:
a ∈ s ⊦ (∀ a0. a0 ∈ s → a0 ≠ a ∧ a0 ∈ s0) ↔ (∀ a. a ∈ s → a ∈ s0) ∧ ¬ a ∈ s
Rule argument
fma: a ∈ s
Additional information
Used formulas: a ∈ s