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Proof step 2 from proof inodes-nodentry (previous step)

ks.inodes?, dentrykey(ino, str) ∈ ks ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
ks.inodes?, dentrykey(ino, str) ∈ ks ⊦ # ks = 0, ks = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ ks ≠ ∅ → # ks ≠ 0; (forward) from specification set-basic[node-index]%ne-01: ⊦ key ∈ ks → ks ≠ ∅; (forward) from specification set-basic[node-index]