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Proof step 12 from proof dups-eqsize-ls (previous step)

mklink(ino, str) ∈ ls, ∀ key. key.dentry? → ¬ mklink(key.ino, key.name) ∈ ls, ∀ ls0, ks. # ls0 < # ls → (∀ key. key ∈ ks → key.dentry?) ∧ (∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls0)) → # ks = # ls0 ⊦ 0 = # ls

User applied rule all left

And closed the goal!

Rule argument

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