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Proof step 2 from proof union-difference (previous step)
| ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s ∪ (s0 \ s) ↔ a ∈ s0 |
User applied rule all left
And closed the goal!
Rule argument
pos: left 1 terms: a abort: F discard: F computed: TAdditional information
Used simplifier rules:
unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;
Used terms: a