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Proof step 6 from proof crash-exists-h (previous step)

⊦ ∃ fs0. (∀ fid0. fid0 ∈ fs0 ↔ fid0 = fid ∧ fid ∈ fs) ∧ (∀ fid. fid ∈ fs0 → fs0[fid] = fs[fid])

User applied rule cut formula

And obtained the following new goals

Proof step 7:
fid ∈ fs ⊦ ∃ fs0. (∀ fid0. fid0 ∈ fs0 ↔ fid0 = fid ∧ fid ∈ fs) ∧ (∀ fid. fid ∈ fs0 → fs0[fid] = fs[fid])
Proof step 12:
⊦ fid ∈ fs, ∃ fs0. (∀ fid0. fid0 ∈ fs0 ↔ fid0 = fid ∧ fid ∈ fs) ∧ (∀ fid. fid ∈ fs0 → fs0[fid] = fs[fid])

Rule argument

fma: fid ∈ fs

Additional information

Used formulas: fid ∈ fs