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Proof step 2 from proof New-same (previous step)

afs-cons(dirs, files), p_ino ∈ (files ++ ino)[ino, af], p_ino ∈ dirs ⊦ ino = 0, ino ∈ dirs, ino ∈ files

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files; (s)%ino ≠ 0, ¬ ino ∈ files ⊦ ino0 ∈ files ++ ino ↔ ino0 = ino ∨ ino0 ∈ files; (s)% ⊦ ino ∈ files[ino0, af] ↔ ino ∈ files; (s)