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Proof step 4 from proof dom-rem (previous step)
| ⊦ a ∈ st \ s ↔ a ∈ dom st \ s |
User applied rule case distinction
And closed the goal!
Rule argument
pos: right 1Additional information
Used simplifier rules:
Diff-ext: ⊦ a ∈ st \ s ↔ a ∈ st ∧ ¬ a ∈ s;%Dom: ⊦ a ∈ dom st ↔ a ∈ st;%Dom: ⊦ a ∈ dom st ↔ a ∈ st;%Diff-ext: ⊦ a ∈ st \ s ↔ a ∈ st ∧ ¬ a ∈ s;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; from specification set-union