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Proof step 2 from proof fids-put1 (previous step)

tree0[parent(p)].dir?, p ∈ tree0 ⊦ tree0[p, tree1].fids = (tree0.fids \ tree0[p].fids) ∪ tree1.fids, p = ε

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
p ∈ tree ⊦ tree[p].fids ⊆ tree.fids
Proof step 4:
tree0[parent(p)].dir?, p ∈ tree0 ⊦ tree0[p, tree1].fids = (tree0.fids ∪ tree1.fids) \ tree0[p].fids, p = ε

Rule argument

Additional information

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