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Proof step 6 from proof inos-write (previous step)

n ∈ ri.inos ∪ (key.inode? ⊃ ∅ ++ key.ino;∅) ⊦ #(ri.inos ++ key.ino) = 0, inodekey(n) = key, inodekey(n) ∈ ri

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: inos-node-store-def: ⊦ n ∈ ri.inos ↔ inodekey(n) ∈ ri;%inos-node-store-def: ⊦ n ∈ ri.inos ↔ inodekey(n) ∈ ri;% ⊦ ns ∪ ∅ = ns; (s)%una: ⊦ (ns ∪ ns0) ∪ ns1 = ns ∪ ns0 ∪ ns1; from specification set-union[nat-set]%unc: ⊦ ns ∪ ns0 = ns0 ∪ ns; from specification set-union[nat-set]% ⊦ (ns ++ n) ∪ ns0 = (ns ∪ ns0) ++ n; (s)% ⊦ n ∈ ns ++ n0 ↔ n = n0 ∨ n ∈ ns; (s)% ⊦ key.inode? → inodekey(key.ino) = key; (s)