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Proof step 17 from proof ofh-fids-new (previous step)

fid = fid0 ∨ n ∈ ofh ∧ ofh[n].fid = fid ⊦ fd ∈ ofh, ∃ n. n ∈ ofh[fd, mkfh(fid0, mode, m)] ∧ ofh[fd, mkfh(fid0, mode, m)][n].fid = fid

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
ofh[n] = mkfh(fid1, mode0, m0), fid = fid0 ∨ n ∈ ofh ∧ fid1 = fid ⊦ fd ∈ ofh, ∃ n. n ∈ ofh[fd, mkfh(fid0, mode, m)] ∧ ofh[fd, mkfh(fid0, mode, m)][n].fid = fid

Rule argument

spec: file-handle name: elim seq: ⊦ (fid = fh.fid ∧ mode = fh.mode) ∧ m = fh.pos ↔ fh = mkfh(fid, mode, m) subst: [fh, fid, mode, m] → [ofh[n], fid1, mode0, m0]

Additional information

Used simplifier rules: ⊦ (fid = fh.fid ∧ mode = fh.mode) ∧ m = fh.pos ↔ fh = mkfh(fid, mode, m); (elim) Used terms: ofh[n], fid1, mode0, m0