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Proof step 4 from proof ex2-shift (previous step)

∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ (_∃ rrP)(H0) ∧ P(H1) ⊦ ∃ r0, r1. ((rrP(r0, r1) * P))(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
P(H0), rrP(r, r0)(H1), H1 ⊥ H0 ⊦ ∃ r0, r1. ((rrP(r0, r1) * P))(H1 ∪ H0)

Rule argument

Additional information

Used simplifier rules: ex2: ⊦ (_∃ rrP)(H) ↔ (∃ r0, r1. rrP(r0, r1)(H));