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Proof step 9 from proof abs-dirs-trunc (previous step)

fs[ri[inodekey(ino)]].directory, inodekey(ino) ∈ ri, dentrykey(ino, str) ∈ ri ⊦ fs[adr, truncnode(key, n)][ri[dentrykey(ino, str)]].ino = fs[ri[dentrykey(ino, str)]].ino, ri[inodekey(ino)] = adr, adr ∈ ri

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: ri[dentrykey(ino, str)] = adr

Additional information

Used simplifier rules: ⊦ adr ≠ adr0 → fs[adr0, nd][adr] = fs[adr]; (s)% ⊦ fs[adr, nd][adr] = nd; (s)%key ∈ ri ⊦ ri[key] ∈ ri; (s) Used formulas: ri[dentrykey(ino, str)] = adr