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Proof step 5 from proof del-addeq (previous step)

a0 ≠ b ∧ a0 ∈ s, ∀ a0. a0 ≠ b ∧ (a0 = a ∨ a0 ∈ s) ↔ a0 = a ∨ a0 ∈ s0 ⊦ a0 ∈ s0, a = b, a ∈ s0, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ s, a1 ∈ s, ∀ a0. a0 ≠ b ∧ (a0 = a ∨ a0 ∈ s) ↔ a0 = a ∨ a0 ∈ s0 ⊦ a0 = b, a = b, a ∈ s, a ∈ s0, a0 ∈ s0

Rule argument

Additional information

Used simplifier rules: In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic