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Proof step 3 from proof int-inse (previous step)

∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0 ++ a ⊦ (¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a ∈ s, ∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0 ++ a ⊦ (¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s
Proof step 6:
∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0 ++ a ⊦ a ∈ s, (¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s

Rule argument

fma: a ∈ s

Additional information

Used formulas: a ∈ s