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Proof step 12 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, ∃ n0. n0 ∈ ofh ∧ ofh[n0].fid = fid0

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 13:
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

Rule argument

fma: fd ∈ ofh \ nats

Additional information

Used simplifier rules: ⊦ fd ∈ ofh \ nats ↔ fd ∈ ofh ∧ ¬ fd ∈ nats; (ws) Used formulas: fd ∈ ofh \ nats