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Proof step 16 from proof mod-other-both (previous step)

rs1 ∩ rs0 = ∅, P(mk(rs1, f)), R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a ⊦ g(r) = a, (a[P] * true)[rs1 ∪ rs0, f(r; o), g(r; a0)], a[P](mk(rs1, f(r; o), g(r; a0)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)
Proof step 18:
rs1 ∩ rs0 = ∅, P(mk(rs1, f)), R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a ⊦ g(r) = a, (a[P] * true)[rs1 ∪ rs0, f(r; o), g(r; a0)], P(mk(rs1, f(r; o))) ∧ (∀ r0. r0 ∈ rs1 → g(r; a0)(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(r; a0), f(r; o), rs1, P, a] flags: <all paths>

Additional information

Used terms: g(r; a0), f(r; o), rs1, P, a