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Proof step 12 from proof restrict-intersect-empty (previous step)

a ∈ s, ∀ a0. ¬ a0 ∈ dom st ∩ s ⊦ # s = 0, s = ∅, (a ∈ st \ s ↔ a ∈ st) ∧ (∅)[a] = st[a]

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a abort: F discard: T computed: T

Additional information

Used simplifier rules: Dom: ⊦ a ∈ dom st ↔ a ∈ st;%Diff-ext: ⊦ a ∈ st \ s ↔ a ∈ st ∧ ¬ a ∈ s;% ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2; (ws)%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union%int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union% ⊦ ¬ a ∈ st → st[a] = (∅)[a]; (s) Used terms: a