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Proof step 32 from proof crash-exists-h (previous step)

str0 ∈ sdt0[str, tree], str0 ∈ sdt0, ∀ fs. ∃ fs0. (∀ fid. fid ∈ fs0 ↔ fid ∈ sdt0[str0].fids ∧ fid ∈ fs) ∧ (∀ fid. fid ∈ fs0 → fs0[fid] = fs[fid]), ∀ str0. str0 ∈ sdt0[str, tree] → (∀ fs0. ∃ fs1. (∀ fid. fid ∈ fs1 ↔ fid ∈ sdt0[str, tree][str0].fids ∧ fid ∈ fs0) ∧ (∀ fid. fid ∈ fs1 → fs1[fid] = fs0[fid])), ∃ fs0. (∀ fid. fid ∈ fs0 ↔ fid ∈ sdt0[str0].fids ∧ fid ∈ fs1) ∧ (∀ fid. fid ∈ fs0 → fs0[fid] = fs1[fid]), ∀ sdt1, fs0. # sdt1 < # sdt0[str, tree] → (∀ str. str ∈ sdt1 → (∀ fs0. ∃ fs1. (∀ fid. fid ∈ fs1 ↔ fid ∈ sdt1[str].fids ∧ fid ∈ fs0) ∧ (∀ fid. fid ∈ fs1 → fs1[fid] = fs0[fid]))) → (∃ fs1. (∀ fid0. fid0 ∈ fs1 ↔ fid0 ∈ sdt1.fids ∧ fid0 ∈ fs0) ∧ (∀ fid. fid ∈ fs1 → fs1[fid] = fs0[fid])) ⊦ str ∈ sdt0, ∃ fs. (∀ fid. fid ∈ fs ↔ fid ∈ sdt0[str0].fids ∧ fid ∈ fs1) ∧ (∀ fid. fid ∈ fs → fs[fid] = fs1[fid])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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