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Proof step 29 from proof synced-dirs-variant1 (previous step)

icache[ino] = mkientry(boolvar0, inode), pcache0[datakey(ino, n)] = mkpentry(true, buf), icache0[ino] = mkientry(boolvar1, inode0), datakey(ino, n) ∈ pcache0, ino ∈ dirs0, ino ∈ icache0 → ¬ boolvar1, ino ∈ dirs → (ino ∈ icache → ¬ boolvar0) ∧ (∀ n. datakey(ino, n) ∈ pcache → ¬ pcache[datakey(ino, n)].dirty), ∀ ino. ino ∈ dirs0 → (ino ∈ icache0 → ¬ icache0[ino].dirty) ∧ (∀ n. datakey(ino, n) ∈ pcache0 → ¬ pcache0[datakey(ino, n)].dirty), ∀ ino. ino ∈ dirs → (ino ∈ icache → ¬ icache[ino].dirty) ∧ (∀ n. datakey(ino, n) ∈ pcache → ¬ pcache[datakey(ino, n)].dirty), ∀ n. datakey(ino, n) ∈ pcache0 → ¬ pcache0[datakey(ino, n)].dirty ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: n abort: F discard: F computed: T

Additional information

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