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Proof step 14 from proof sep (previous step)

mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)) ⊦ rs0 ∪ rs1 = rs ∧ P0(mk(rs0, f)) ∧ P1(mk(rs1, f))

User applied rule case distinction

And obtained the following new goals

Proof step 15:
mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)) ⊦ rs0 ∪ rs1 = rs
Proof step 22:
mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)) ⊦ P0(mk(rs0, f))
Proof step 30:
mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)) ⊦ P1(mk(rs1, f))

Rule argument

pos: right 1

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