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Proof step 20 from proof fids-del (previous step)

(tree -- p0).fids = tree.fids \ tree[p0].fids, p0 ∈ tree ⊦ p0 = ε, sdt0.fids ∪ (tree -- p0).fids = (sdt0.fids ∪ tree.fids) \ tree[p0].fids, str ∈ sdt0

User applied rule insert equation

And obtained the following new goals

Proof step 21:
p ∈ tree ⊦ tree[p].fids ⊆ tree.fids

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: fids-sub: p ∈ tree ⊦ tree[p].fids ⊆ tree.fids;%fidb ⊆ fidb0 ⊦ (fidb0 \ fidb) ∪ fidb1 = (fidb0 ∪ fidb1) \ fidb; (s)%union-ass: ⊦ fidb ∪ fidb0 ∪ fidb1 = (fidb ∪ fidb0) ∪ fidb1; from specification bag-union[fidbag]%union-com: ⊦ fidb ∪ fidb0 = fidb0 ∪ fidb; from specification bag-union[fidbag]