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Proof step 4 from proof union-sizele (previous step)

#(s ∪ s0) = # s, #(s ∪ s0) ≤ # s ⊦ ∀ a. a ∈ s0 → a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ s0, a0 ∈ s0, #(s ∪ s0) ≤ # s, ∀ a. a ∈ s0 → a ∈ s ⊦ a ∈ s

Rule argument

Additional information

Used simplifier rules: union-sizeeq: ⊦ #(s ∪ s0) = # s ↔ s0 ⊆ s;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic% ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); (s)