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Proof step 20 from proof lift-unknown (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], r ==>(o, a)[rs1, f, g] ⊦ ∃ o. (a[r => o] * R)[rs1 ∪ rs0, f, g], (∃ o. ((r => o))(mk(rs1, f))) ∧ (r0 ∈ rs1 → g(r0) = a)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ R[rs, f, g] ↔ R(mk(rs, f, g))
Proof step 22:
rs1 ∩ rs0 = ∅, r ==>(o, a)(mk(rs1, f, g)), R[rs0, f, g] ⊦ ∃ o. (a[r => o] * R)[rs1 ∪ rs0, f, g], (∃ o. ((r => o))(mk(rs1, f))) ∧ (r0 ∈ rs1 → g(r0) = a)

Rule argument

spec: heap-fun, inst: heap-aux name: apply seq: ⊦ R[rs, f, g] ↔ R(mk(rs, f, g)) subst: [g, f, rs, R] → [g, f, rs1, r ==>(o, a)] flags: <all paths>

Additional information

Used terms: g, f, rs1, r ==>(o, a)