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Proof step 20 from proof synced-dirs-add-new-clean-inode (previous step)

pcache[datakey(ino0, n)] = mkpentry(true, buf), icache[ino0] = mkientry(boolvar0, inode0), datakey(ino0, n) ∈ pcache, ino0 ∈ dirs, ino ∈ dirs → (∀ n. datakey(ino, n) ∈ pcache → ¬ pcache[datakey(ino, n)].dirty), ino0 ∈ icache → ¬ boolvar0, ∀ n. datakey(ino0, n) ∈ pcache → ¬ pcache[datakey(ino0, n)].dirty, ∀ ino0. ino0 ∈ dirs → (ino0 ∈ icache[ino, mkientry(false, inode)] → ¬ icache[ino, mkientry(false, inode)][ino0].dirty) ∧ (∀ n. datakey(ino0, n) ∈ pcache → ¬ pcache[datakey(ino0, n)].dirty) ⊦ ino0 = ino, ino ∈ icache

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ mkpentry(boolvar, buf).dirty ↔ boolvar; (s) Used terms: n