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Proof step 7 from proof del-addeq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a0 ∈ s0, a1 ∈ s0, ∀ a0. a0 ≠ b ∧ (a0 = a ∨ a0 ∈ s) ↔ a0 = a ∨ a0 ∈ s0 ⊦ a = b, a ∈ s, a ∈ s0, a0 ≠ b ∧ a0 ∈ s

Rule argument

Additional information

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