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Proof step 27 from proof aux-known (previous step)

r0 ∈ rs1 ↔ r0 = r, rs1 ∩ rs0 = ∅, R[rs0, f, g], r0 = r → g(r) = a0, r0 = r → f(r) = o, ∀ r1. (r1 ∈ rs1 ↔ r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ o;(∅ ++ r)[r1])), ∀ r. r ∈ rs1 → g(r) = a0 ⊦ r = null, R[rs0, f, g(r; a1)], (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 28:
⊦ ¬ r ∈ rs → (R[rs, f, g(r; a)] ↔ R[rs, f, g])
Proof step 29:
r0 ∈ rs1 ↔ r0 = r, rs1 ∩ rs0 = ∅, R[rs0, f, g], r ∈ rs0, r0 = r → f(r) = o, r0 = r → g(r) = a0, ∀ r. r ∈ rs1 → g(r) = a0, ∀ r1. (r1 ∈ rs1 ↔ r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ o;(∅ ++ r)[r1])) ⊦ r = null, R[rs0, f, g(r; a1)], (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0

Rule argument

spec: <toplevel> name: switch-aux seq: ⊦ ¬ r ∈ rs → (R[rs, f, g(r; a)] ↔ R[rs, f, g]) subst: [a, r, g, f, rs, R] → [a1, r, g, f, rs0, R] flags: <all paths>

Additional information

Used terms: a1, r, g, f, rs0, R