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Proof step 26 from proof dups-eqsize-ls (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ alink ∈ ls ↔ (∃ ls0. ls = ls0 ++ alink ∧ ¬ alink ∈ ls0)
Proof step 28:
∀ key0. key0 ∈ ks0 → key0.dentry?, ∀ key0. key0.dentry? → (key0 = dentrykey(m, str) ∨ key0 ∈ ks0 ↔ mklink(key0.ino, key0.name) ∈ ls), ∃ ls0. ls = ls0 ++ mklink(m, str) ∧ ¬ mklink(m, str) ∈ ls0, ∀ 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

spec: set-basic, inst: alink-set name: indef seq: ⊦ alink ∈ ls ↔ (∃ ls0. ls = ls0 ++ alink ∧ ¬ alink ∈ ls0) subst: [ls, alink] → [ls, mklink(m, str)] flags: <all paths>

Additional information

Used terms: ls, mklink(m, str)