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Proof step 35 from proof initialization (previous step)

⌊err⌋ ⊦ ∃ root, fs, ofh, psynced. ⟨choose err with ⌊err⌋ in err0 := err⟩ (tree = root ∧ fs1 = fs ∧ ofh1 = ofh ∧ (isdir ↔ psynced) ∧ err = err0)

User applied rule exists right

And obtained the following new goals

Proof step 36:
⌊err⌋ ⊦ ⟨psynced := isdir, ofh := ofh1, fs := fs1, root := tree; choose err with ⌊err⌋ in err0 := err⟩ (tree = root ∧ fs1 = fs ∧ ofh1 = ofh ∧ (isdir ↔ psynced) ∧ err = err0)

Rule argument

pos: right 1 terms: tree, fs1, ofh1, isdir abort: F discard: T computed: F

Additional information

Used terms: tree, fs1, ofh1, isdir