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Proof step 6 from proof sep (previous step)

mk(rs, f) = H0 ∪ H, P0(H0), P1(H), H0 ⊥ H ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0(mk(rs0, f)) ∧ P1(mk(rs1, f))

User applied rule apply lemma

And obtained the following new goals

Proof step 7:
⊦ ∃ rs, f. H = mk(rs, f)
Proof step 8:
mk(rs, f) = H0 ∪ H, P0(H0), P1(H), H0 ⊥ H, ∃ rs, f. H0 = mk(rs, f) ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0(mk(rs0, f)) ∧ P1(mk(rs1, f))

Rule argument

spec: <toplevel> name: mk-exists seq: ⊦ ∃ rs, f. H = mk(rs, f) subst: [H] → [H0]

Additional information

Used terms: H0