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Proof step 2 from proof synced-dirs-variant1 (previous step)

synced(dirs, icache, pcache) ⊦ ¬ synced(dirs0, icache0, pcache0) ↔ ¬(∀ ino. (ino ∈ dirs → synced(ino, icache, pcache)) → ino ∈ dirs0 → synced(ino, icache0, pcache0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ ino. ino ∈ dirs → (ino ∈ icache → ¬ icache[ino].dirty) ∧ (∀ 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)) → ino ∈ dirs0 → (ino ∈ icache0 → ¬ icache0[ino].dirty) ∧ (∀ n. datakey(ino, n) ∈ pcache0 → ¬ pcache0[datakey(ino, n)].dirty))

Rule argument

Additional information

Used simplifier rules: synced-dirs: ⊦ synced(dirs, icache, pcache) ↔ (∀ ino. ino ∈ dirs → synced(ino, icache, pcache));%synced-ino: ⊦ synced(ino, icache, pcache) ↔ (ino ∈ icache → ¬ icache[ino].dirty) ∧ (∀ n. datakey(ino, n) ∈ pcache → ¬ pcache[datakey(ino, n)].dirty);