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Proof step 11 from proof fids-sub (previous step)

p0 ∈ tree, ∀ tree. p0 ∈ tree → tree[p0].fids ⊆ tree.fids ⊦ str ∈ sdt0, tree[p0].fids ⊆ sdt0.fids ∪ tree.fids

User applied rule all left

And obtained the following new goals

Proof step 12:
p0 ∈ tree, p0 ∈ tree → tree[p0].fids ⊆ tree.fids ⊦ str ∈ sdt0, tree[p0].fids ⊆ sdt0.fids ∪ tree.fids

Rule argument

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

Additional information

Used terms: tree