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Proof step 13 from proof ofh-fids-new (previous step)

ofh[fd, mkfh(fid, mode, m)][n] = mkfh(fid0, mode0, m0), n ∈ ofh[fd, mkfh(fid, mode, m)] ⊦ fid0 = fid, fd ∈ ofh, fd ∈ ofh \ nats, ∃ n0. n0 ∈ ofh ∧ ofh[n0].fid = fid0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: n abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ fd ∈ ofh[fd0, fh] ↔ fd = fd0 ∨ fd ∈ ofh; (ws)% ⊦ fd = fd0 → ofh[fd0, fh][fd] = fh; (s)%⊦ mkfh(fid, mode, m) = mkfh(fid0, mode0, m0) ↔ fid = fid0 ∧ mode = mode0 ∧ m = m0; (ws)% ⊦ fd ∈ ofh ∧ ¬ fd0 ∈ ofh → ofh[fd0, fh][fd] = ofh[fd]; (s)% ⊦ mkfh(fid, mode, m).fid = fid; (s) Used terms: n