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Proof step 1 from proof del-feq-01 (conclusion)

⊦ (s ++ a) -- b = s0 ++ c ∧ ¬ c ∈ s → a = c ∧ b ≠ c

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-feq: ⊦ s -- a = s0 ++ b → a ≠ b ∧ b ∈ s ∧ ¬ a ∈ s0;%in-03: ⊦ ¬ a ∈ s → (a ∈ s ++ b ↔ a = b); from specification set-basic