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Proof step 2 from proof union-difference (previous step)

s ⊆ s0 ⊦ s ∪ (s0 \ s) = s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ s, a0 ∈ s, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s0

Rule argument

Additional information

Used simplifier rules: diff-union-01: ⊦ ¬ a ∈ s → (a ∈ (s \ s0) ∪ s1 ↔ a ∈ s1);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%unrax: ⊦ a ∈ s ∪ s0 ↔ a ∈ s ∨ a ∈ s0;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;% ⊦ ¬ a ∈ ∅; (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic% ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)