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Proof step 135 from proof replayone-prog (previous step)

ri0 = ri, fs[adr].trunc?, node-cons(fs[adr].key, fs[adr]), adr ∈ fs, ¬ fs[adr].data?, ¬ fs[adr].dentry?, ¬ fs[adr].inode?, ¬ adr ∈ ri ⊦ ⟪aubifs_core_index_truncate#(fs[adr].key, fs[adr].size, fi, ro, fo, fs, log, boolvar0; ri)⟫ (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-04 seq: ⊦ nd.trunc? → (key = nd.key ∧ m = nd.size ↔ nd = truncnode(key, m)) subst: [nd, key, m] → [fs[adr], key, m]

Additional information

Used simplifier rules: ⊦ nd.trunc? → (key = nd.key ∧ m = nd.size ↔ nd = truncnode(key, m)); (elim) Used terms: fs[adr], key, m