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Proof step 6 from proof union-sizeeq (previous step)

s ⊆ s ∪ s0 ∧ # s = #(s ∪ s0) → s = s ∪ s0, #(s ∪ s0) = # s, a ∈ s0, a0 ∈ s0 ⊦ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
#(s ∪ s0) = # s, a ∈ s0, a0 ∈ s0, ∀ a1. a1 ∈ s0 → a1 ∈ s ⊦ a ∈ s

Rule argument

Additional information

Used simplifier rules: unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic%Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); from specification set-basic