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Proof step 19 from proof write-other-true (previous step)

rs1 ∩ rs0 = ∅, a[P](mk(rs1, f(r0; o), g)), r0 ∈ rs1 ⊦ # rs1 = 0, g(r0) = a, o = f(r0), rs1 = ∅, (a[P] * true)[rs1 ∪ rs0, f, g], a[P](mk(rs1, f, g))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)
Proof step 21:
rs1 ∩ rs0 = ∅, P(mk(rs1, f(r0; o))), r0 ∈ rs1, ∀ r. r ∈ rs1 → g(r) = a ⊦ # rs1 = 0, g(r0) = a, o = f(r0), rs1 = ∅, a[P](mk(rs1, f, g)), (a[P] * true)[rs1 ∪ rs0, f, g]

Rule argument

spec: <toplevel> 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] → [g, f(r0; o), rs1, P, a] flags: <all paths>

Additional information

Used terms: g, f(r0; o), rs1, P, a