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Proof step 6 from proof afs-cons-nlink0-notin (previous step)

afs-cons(dirs, files), alink.pino ∈ dirs, alink.name ∈ dirs[alink.pino].entries ⊦ # links(dirs[alink.pino].entries[alink.name], dirs) = 0, links(dirs[alink.pino].entries[alink.name], dirs) = ∅, dirs[alink.pino].entries[alink.name] ∈ files, dirs[alink.pino].entries[alink.name] ∈ dirs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
alink = mklink(m, str), afs-cons(dirs, files), str ∈ dirs[m].entries, m ∈ dirs ⊦ # links(dirs[m].entries[str], dirs) = 0, links(dirs[m].entries[str], dirs) = ∅, dirs[m].entries[str] ∈ dirs, dirs[m].entries[str] ∈ files

Rule argument

spec: alink name: elim seq: ⊦ m = alink.pino ∧ str = alink.name ↔ alink = mklink(m, str) subst: [alink, m, str] → [alink, m, str]

Additional information

Used simplifier rules: ⊦ m = alink.pino ∧ str = alink.name ↔ alink = mklink(m, str); (elim) Used terms: alink, m, str