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Proof step 5 from proof union-subset-02 (previous step)

a0 ∈ s1, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ s1, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s → a ∈ s0 ⊦ s1 = ∅, a0 = a, # s1 = 0, a0 ∈ s2, a0 ∈ s0

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic