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Proof step 12 from proof embed-and (previous step)

⊦ P0(mk(rs, f)) ∧ P1(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a) ↔ a[P0](mk(rs, f, g)) ∧ a[P1](mk(rs, f, g))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)
Proof step 14:
⊦ P0(mk(rs, f)) ∧ P1(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a) ↔ (P0(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)) ∧ a[P1](mk(rs, 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, rs, P0, a] flags: <all paths>

Additional information

Used terms: g, f, rs, P0, a