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Proof step 29 from proof replayone-prog (previous step)

fs[adr].nlink = 0, ri0 = ri, fs[adr].key.ino ≠ ROOT_INO, fs[adr].inode?, node-cons(fs[adr].key, fs[adr]), adr ∈ fs, ¬ fs[adr].directory, ¬ adr ∈ ri ⊦ ⟪aubifs_core_index_truncate#(fs[adr].key, 0, fi, ro, fo, fs, log, boolvar0; ri); nd := fs[adr]; err := ESUCCESS; if err = ESUCCESS then aubifs_core_index_remove#(nd.key, fi, ro, fo, fs, log, boolvar0; ri); if nd.dentry? then if nd.ino ≠ 0 then aubifs_core_index_store#(nd.key, adr, fi, ro, fo, fs, log, boolvar0; ri) else aubifs_core_index_remove#(nd.key, fi, ro, fo, fs, log, boolvar0; ri); boolvar := boolvar0; if nd.data? then {let key = inodekey(nd.key.ino) in aubifs_core_index_contains#(key, ri, fi, ro, fo, fs, log, boolvar0; boolvar; err); if err = ESUCCESS ∧ boolvar then aubifs_core_index_store#(nd.key, adr, fi, ro, fo, fs, log, boolvar0; ri)}; if nd.trunc? then aubifs_core_index_truncate#(nd.key, nd.size, fi, ro, fo, fs, log, boolvar0; ri)⟫ ((err = ESUCCESS ∨ ⌊err⌋ ) ∧ (err = ESUCCESS → ri = replay-one(adr, fs[adr], ri0)))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: node name: elim-03 seq: ⊦ nd.inode? → ( ( ( ( (key = nd.key ∧ md = nd.meta) ∧ (boolvar ↔ nd.directory)) ∧ m = nd.nlink) ∧ m0 = nd.nsubdirs) ∧ m1 = nd.size ↔ nd = inodenode(key, md, boolvar, m, m0, m1)) subst: [nd, key, md, boolvar, m, m0, m1] → [fs[adr], key0, md, boolvar1, m, m0, m1]

Additional information

Used simplifier rules: ⊦ nd.inode? → ( ( ( ( (key = nd.key ∧ md = nd.meta) ∧ (boolvar ↔ nd.directory)) ∧ m = nd.nlink) ∧ m0 = nd.nsubdirs) ∧ m1 = nd.size ↔ nd = inodenode(key, md, boolvar, m, m0, m1)); (elim) Used terms: fs[adr], key0, md, boolvar1, m, m0, m1