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Proof step 3 from proof diff-del (previous step)

a0 ∈ s1 \ s2 ⊦ a0 ∈ (s1 -- a) \ (s2 -- a), a ∈ s1 \ s2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a0 ∈ s1, a1 ∈ s1 ⊦ a0 ∈ s2, a ∈ s1 \ s2, a0 ∈ (s1 -- a) \ (s2 -- a)

Rule argument

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%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