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Proof step 3 from proof k-intersect-rec2 (previous step)

¬ a ∈ s ∨ ¬ a ∈ s0 ⊦ a0 ∈ s ∧ a0 ∈ s0 ↔ a0 ∈ s -- a ∧ a0 ∈ s0 -- a

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a ∈ s0

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%del-in: ⊦ ¬ a ∈ s → s -- a = s; Used formulas: a ∈ s0