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Proof step 1 from proof sep (conclusion)

⊦ (P0 * P1)[rs, f] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0[rs0, f] ∧ P1[rs1, f])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ((P0 * P1))(mk(rs, f)) ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0(mk(rs0, f)) ∧ P1(mk(rs1, f)))

Rule argument

Additional information

Used simplifier rules: apply: ⊦ P[rs, f] ↔ P(mk(rs, f));