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Proof step 6 from proof Diff-ext (
previous step
)
∀ a, st. a ∈ st \ s0 ↔ a ∈ st ∧ ¬ a ∈ s0 ⊦ a ∈ st \ s0 ↔ a ∈ st ∧ ¬ a ∈ s0, a = a0
User applied rule all left
And closed the goal!
Rule argument
pos: left 1 terms: a, st abort: F discard: T computed: T
Additional information
Used terms: a, st