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Proof step 6 from proof elems-in-notin-ginval (previous step)

n ∈ ns, n ∈ nx, ∀ n0. ¬ n0 ∈ elems nx ∩ ns ⊦ # ns = 0, ns = ∅

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n abort: F discard: T computed: F

Additional information

Used simplifier rules: elems-def: ⊦ n ∈ elems nx ↔ n ∈ nx;% ⊦ n ∈ ns ∩ ns0 ↔ n ∈ ns ∧ n ∈ ns0; (ws)%int-a: ⊦ (ns ∩ ns0) ∩ ns1 = ns ∩ ns0 ∩ ns1; from specification set-union[nat-set]%int-c: ⊦ ns ∩ ns0 = ns0 ∩ ns; from specification set-union[nat-set] Used terms: n