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Proof step 53 from proof aux-known (previous step)

rs1 ∩ rs0 = ∅, a1[r => o][rs1, f, g(r; a1)], R[rs0, f, g(r; a1)] ⊦ (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], g(r)[r => o][rs1, f, g] ∧ R[rs0, f, g]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 54:
⊦ ¬ r ∈ rs → (R[rs, f, g(r; a)] ↔ R[rs, f, g])
Proof step 55:
rs1 ∩ rs0 = ∅, R[rs0, f, g(r; a1)], a1[r => o][rs1, f, g(r; a1)], r ∈ rs0 ⊦ (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], g(r)[r => o][rs1, f, g] ∧ R[rs0, f, g]
Proof step 68:
rs1 ∩ rs0 = ∅, R[rs0, f, g], a1[r => o][rs1, f, g(r; a1)] ⊦ (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], r ∈ rs0, g(r)[r => o][rs1, f, g] ∧ R[rs0, f, g]

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