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Proof step 7 from proof Diff-get (
previous step
)
∀ st, a. (st \ s0)[a] = st[a] ∨ a ∈ s0 ⊦ (st \ s0)[a] = st[a], a = a0, a ∈ s0 ++ a0
User applied rule all left
And closed the goal!
Rule argument
pos: left 1 terms: st, a abort: F discard: T computed: T
Additional information
Used simplifier rules: ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s); (s) Used terms: st, a