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Proof step 4 from proof mapex-and (previous step)

⊦ (_∃(λ o. r => o) ∧ r => o) = r => o, ∀ H. ((r => o → _∃(λ o. r => o)))(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ (_∃(λ o. r => o) ∧ r => o) = r => o, r = null, ∃ o0. (∅ ++ r)[r, o] = (∅ ++ r)[r, o0]

Rule argument

Additional information

Used simplifier rules: maplet: ⊦ ((r => o))(H) ↔ H = (∅ ++ r)[r, o] ∧ r ≠ null;%ex3: ⊦ (_∃ oP)(H) ↔ (∃ o. oP(o)(H));% ⊦ ((P0 → P1))(H) ↔ P0(H) → P1(H); (s)