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Proof step 25 from proof dups-eqsize-ls (previous step)

key = dentrykey(m, str), key.dentry?, mklink(m, str) ∈ ls, ∀ key0. key0 ∈ ks0 → key0.dentry?, ∀ key0. key0.dentry? → (key0 = key ∨ key0 ∈ ks0 ↔ mklink(key0.ino, key0.name) ∈ ls), ∀ ls0, ks. # ls0 < # ls → (∀ key. key ∈ ks → key.dentry?) ∧ (∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls0)) → # ks = # ls0 ⊦ # ls = 0, # ks0 + 1 = # ls, ls = ∅, key ∈ ks0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
mklink(m, str) ∈ ls, ∀ key0. key0.dentry? → (key0 = dentrykey(m, str) ∨ key0 ∈ ks0 ↔ mklink(key0.ino, key0.name) ∈ ls), ∀ key0. key0 ∈ ks0 → key0.dentry?, ∀ ls0, ks. # ls0 < # ls → (∀ key. key ∈ ks → key.dentry?) ∧ (∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls0)) → # ks = # ls0 ⊦ # ls = 0, # ks0 + 1 = # ls, ls = ∅, dentrykey(m, str) ∈ ks0

Rule argument

Additional information

Used simplifier rules: ⊦ dentrykey(m, str).dentry?; (s)