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Proof step 10 from proof merge-owned (previous step)

rs1 ∩ rs0 = ∅, rs2 ∩ rs = ∅, rs2 ∪ rs = rs1 ∪ rs0, R[rs0, f, g], a[P][rs1, f, g], R0[rs, f, g], a0[P0][rs2, f, g], rs2 ⊆ rs0 ⊦ a0 = a, (a[P] * a0[P0] * true)[rs1 ∪ rs0, f, g]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ rs ⊆ rs0 ↔ (∃ rs1. rs ∪ rs1 = rs0 ∧ rs ∩ rs1 = ∅)
Proof step 12:
rs2 ∩ rs = ∅, rs1 ∩ rs0 = ∅, rs2 ∪ rs = rs1 ∪ rs0, a0[P0][rs2, f, g], R0[rs, f, g], a[P][rs1, f, g], R[rs0, f, g], ∃ rs1. rs2 ∪ rs1 = rs0 ∧ rs2 ∩ rs1 = ∅ ⊦ a0 = a, (a[P] * a0[P0] * true)[rs1 ∪ rs0, f, g]

Rule argument

spec: set-union, inst: refset name: subset-is-union seq: ⊦ rs ⊆ rs0 ↔ (∃ rs1. rs ∪ rs1 = rs0 ∧ rs ∩ rs1 = ∅) subst: [rs0, rs] → [rs0, rs2] flags: <all paths>

Additional information

Used terms: rs0, rs2