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Proof step 1 from proof del-feq-02 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ a0. a0 ≠ b ∧ (a0 = a ∨ a0 ∈ s) ↔ a0 = c ∨ a0 ∈ s0 ⊦ a = c, a = b, a ∈ s0

Rule argument

Additional information

Used simplifier rules: del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic