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Proof step 11 from proof ofh-fids-new (previous step)

n ∈ ofh[fd, mkfh(fid, mode, m)] ⊦ ofh[fd, mkfh(fid, mode, m)][n].fid = fid, fd ∈ ofh, ∃ n0. n0 ∈ ofh ∧ ofh[n0].fid = ofh[fd, mkfh(fid, mode, m)][n].fid

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
ofh[fd, mkfh(fid, mode, m)][n] = mkfh(fid0, mode0, m0), n ∈ ofh[fd, mkfh(fid, mode, m)] ⊦ fid0 = fid, fd ∈ ofh, ∃ n0. n0 ∈ ofh ∧ ofh[n0].fid = fid0

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[fd, mkfh(fid, mode, m)][n], fid0, 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[fd, mkfh(fid, mode, m)][n], fid0, mode0, m0