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Proof step 107 from proof write-known (previous step)

rs1 ∩ rs0 = ∅, ((r => o1))(mk(rs1, f(r; o1))), R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a ⊦ (a[r => f(r)] * R)[rs1 ∪ rs0, f, g], a[r => f(r)](mk(rs1, f, g)), r ∈ rs0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 108:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)
Proof step 109:
rs1 ∩ rs0 = ∅, ((r => o1))(mk(rs1, f(r; o1))), R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a ⊦ (a[r => f(r)] * R)[rs1 ∪ rs0, f, g], r ∈ rs0, ((r => f(r)))(mk(rs1, f)) ∧ (∀ r. r ∈ rs1 → g(r) = a)

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, rs1, r => f(r), a] flags: <all paths>

Additional information

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