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Proof step 18 from proof ofh-fids-new (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 19:
ofh[n] = mkfh(fid1, mode0, m0), fid = fid0 ⊦ fd ∈ ofh, ∃ n. n ∈ ofh[fd, mkfh(fid0, mode, m)] ∧ ofh[fd, mkfh(fid0, mode, m)][n].fid = fid
Proof step 21:
ofh[n] = mkfh(fid1, mode0, m0), fid1 = fid, n ∈ ofh ⊦ fd ∈ ofh, ∃ n. n ∈ ofh[fd, mkfh(fid0, mode, m)] ∧ ofh[fd, mkfh(fid0, mode, m)][n].fid = fid

Rule argument

pos: left 2

Proof step 18 from proof ofh-fids-new (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information