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Proof step 30 from proof links-add-entry (previous step)

p_ino ∈ dirs, alink.pino ∈ dirs, ¬ str ∈ entries ∨ entries[str] ≠ ino0 ⊦ alink.name ∈ (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries[str, ino]);dirs[alink.pino]).entries ∧ ino0 = (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries[str, ino]);dirs[alink.pino]).entries[alink.name] ↔ alink.name ∈ (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries);dirs[alink.pino]).entries ∧ ino0 = (alink.pino = p_ino ⊃ mkdir(m0, md, n, n0, entries);dirs[alink.pino]).entries[alink.name], ino = ino0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 31:
alink = mklink(m, str0), m ∈ dirs, p_ino ∈ dirs, ¬ str ∈ entries ∨ entries[str] ≠ ino0 ⊦ str0 ∈ (m = p_ino ⊃ mkdir(m0, md, n, n0, entries[str, ino]);dirs[m]).entries ∧ ino0 = (m = p_ino ⊃ mkdir(m0, md, n, n0, entries[str, ino]);dirs[m]).entries[str0] ↔ str0 ∈ (m = p_ino ⊃ mkdir(m0, md, n, n0, entries);dirs[m]).entries ∧ ino0 = (m = p_ino ⊃ mkdir(m0, md, n, n0, entries);dirs[m]).entries[str0], ino = ino0

Rule argument

spec: alink name: elim seq: ⊦ m = alink.pino ∧ str = alink.name ↔ alink = mklink(m, str) subst: [alink, m, str] → [alink, m, str0]

Additional information

Used simplifier rules: ⊦ m = alink.pino ∧ str = alink.name ↔ alink = mklink(m, str); (elim) Used terms: alink, m, str0