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Proof step 12 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. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0(mk(rs0, f)) ∧ P1(mk(rs1, f))

User applied rule exists right

And obtained the following new goals

Proof step 13:
mk(rs, f) = mk(rs0, f0) ∪ mk(rs1, f1), rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)) ⊦ rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0(mk(rs0, f)) ∧ P1(mk(rs1, f))

Rule argument

pos: right 1 terms: rs0, rs1 abort: F discard: T computed: F

Additional information

Used terms: rs0, rs1