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Proof step 59 from proof rely-owned-unchgd (previous step)

rs2 ∩ rs1 = ∅, a[P](mk(rs2, f0, g0)), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs2 ∪ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ rs2 ∪ (rs \ rs2) = rs, rs2 ⊆ rs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 60:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)
Proof step 61:
rs2 ∩ rs1 = ∅, P(mk(rs2, f0)), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs2 ∪ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)), ∀ r. r ∈ rs2 → g0(r) = a ⊦ rs2 ∪ (rs \ rs2) = rs, rs2 ⊆ rs

Rule argument

spec: embed-aux name: lift-aux seq: ⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a) subst: [g, f, rs, P, a] → [g0, f0, rs2, P, a] flags: <all paths>

Additional information

Used terms: g0, f0, rs2, P, a