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Proof step 18 from proof inv-seek (previous step)

err = ESUCCESS, err0 = err1, f = fs[fh.fid], fh = ofh[fd], whence = SEEK_END, root.dir?, fids-cons(root, fs, ofh), pre-seek(fd, n, whence, err, user, root, fs, ofh), fh.fid ∈ fs, fd ∈ ofh ⊦ whence = SEEK_CUR, fids-cons(root, fs, ofh[fd, ofh[fd].pos:= (# f.content + n)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
⊦ fids-cons(tree0, fs0, ofh0) → (fids-cons(tree1, fs1, ofh1) ↔ (∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids)))
Proof step 20:
fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids, root.dir?, fids-cons(root, fs, ofh), ofh[fd].fid ∈ fs, fd ∈ ofh ⊦ fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh[fd, mkfh(ofh[fd].fid, ofh[fd].mode, # fs[ofh[fd].fid].content + n)].fids

Rule argument

Additional information

Used simplifier rules: fids-cons-reduce: ⊦ fids-cons(tree0, fs0, ofh0) → ( fids-cons(tree1, fs1, ofh1) ↔ (∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids)));% ⊦ fh.pos:= m = mkfh(fh.fid, fh.mode, m); (s)% ⊦ pre-seek(fd, n, whence, ESUCCESS, user, root, fs, ofh) ↔ fd ∈ ofh; (s)% ⊦ SEEK_CUR ≠ SEEK_END; (s)