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Proof step 16 from proof inv-recover (
previous step
)
fid ∈ fs1 ↔ fid ∈ tree.fids ∨ fid ∈ ofh1.fids, tree.dir?, fids-cons(tree, fs1, ofh1), ∀ fid. fid ∈ fs0 → fs0[fid] = fs1[fid], ∀ fid. fid ∈ fs0 ↔ fid ∈ tree.fids ∧ fid ∈ fs1 ⊦ fid ∈ fs0 ↔ fid ∈ tree.fids
User applied rule all left
And closed the goal!
Rule argument
pos: left 5 terms: fid abort: F discard: F computed: T
Additional information
Used terms: fid