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Proof step 10 from proof dups-eqsize-entries (previous step)

str ∈ entries, ∀ key. key.dentry? ∧ key.ino = ino → ¬ key.name ∈ entries, ∀ ks0, ino, entries. # ks0 < # ∅ → (∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks0 ↔ key.name ∈ entries)) → # ks0 = # entries ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: dentrykey(ino, str) abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ dentrykey(m, str).dentry?; (s)% ⊦ dentrykey(m, str).ino = m; (s)% ⊦ dentrykey(m, str).name = str; (s) Used terms: dentrykey(ino, str)