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Proof step 15 from proof aux-known (previous step)

rs1 ∩ rs0 = ∅, a0[r => o](mk(rs1, f, g)), R[rs0, f, g] ⊦ (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0, ((r => o))(mk(rs1, f)) ∧ (r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1) ∧ R[rs0, f, g(r; a1)]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ a[P](mk(rs, f, g)) ↔ P(mk(rs, f)) ∧ (∀ r. r ∈ rs → g(r) = a)
Proof step 17:
rs1 ∩ rs0 = ∅, ((r => o))(mk(rs1, f)), R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a0 ⊦ ((r => o))(mk(rs1, f)) ∧ (r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1) ∧ R[rs0, f, g(r; a1)], (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0

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 => o, a0] flags: <all paths>

Additional information

Used terms: g, f, rs1, r => o, a0