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Proof step 20 from proof lift-ex-r (previous step)

a[rP(r)](mk(rs, f, g)) ⊦ (∃ r. rP(r)(mk(rs, f))) ∧ (r0 ∈ rs → g(r0) = a)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)
Proof step 22:
rP(r)(mk(rs, f)), ∀ r. r ∈ rs → g(r) = a ⊦ (∃ r. rP(r)(mk(rs, f))) ∧ (r0 ∈ rs → g(r0) = 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, rs, rP(r), a] flags: <all paths>

Additional information

Used terms: g, f, rs, rP(r), a