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Proof step 5 from proof diff-union-03 (previous step)

a ∈ s \ s1 ∨ a ∈ s0 \ s1 ⊦ (a ∈ s ∨ a ∈ s0) ∧ ¬ a ∈ s1

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic%In-empty: ⊦ ¬ a ∈ ∅; from specification infset-basic