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Proof step 14 from proof lift-ex-o (
previous step
)
oP(o)(mk(rs, f)), ∀ r. r ∈ rs → g(r) = a ⊦ a[oP(o)](mk(rs, f, g)), ∃ o. a[oP(o)](mk(rs, f, g))
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 15
:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → 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, rs, oP(o), a] flags: <all paths>
Additional information
Used terms: g, f, rs, oP(o), a