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Proof step 50 from proof lift-unknown (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], a[r =>_](mk(rs1, f, g)) ⊦ r ==>_[rs1, f, g], (r ==>_ * R)[rs1 ∪ rs0, f, g]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 51:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)
Proof step 52:
rs1 ∩ rs0 = ∅, r =>_(mk(rs1, f)), R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a ⊦ (r ==>_ * R)[rs1 ∪ rs0, f, g], r ==>_[rs1, 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, rs1, r =>_, a] flags: <all paths>

Additional information

Used terms: g, f, rs1, r =>_, a