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Proof step 1 from proof sub-del-05 (conclusion)

⊦ a ≠ b ∧ ¬ a ∈ s → s -- b ⊆ s -- b -- a

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: subset-del: ⊦ s ⊆ s -- a ↔ ¬ a ∈ s;%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);