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Proof step 66 from proof aux-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g(r; a1)], r ∈ rs0, ∀ r0. r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1, ∀ r0. (r0 ∈ mk(rs1, f) ↔ r0 ∈ (∅ ++ r)[r, o]) ∧ (r0 ∈ mk(rs1, f) → mk(rs1, f)[r0] = (∅ ++ r)[r, o][r0]) ⊦ # rs0 = 0, r = null, rs0 = ∅, (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], g(r)[r => o][rs1, f, g] ∧ R[rs0, f, g]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 67:
rs1 ∩ rs0 = ∅, R[rs0, f, g(r; a1)], r ∈ rs0, ∀ r0. r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1, ∀ r0. (r0 ∈ rs1 ↔ r0 = r) ∧ (r0 ∈ rs1 → f(r0) = (r0 = r ⊃ o;(∅ ++ r)[r0])) ⊦ # rs0 = 0, r = null, rs0 = ∅, (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], g(r)[r => o][rs1, f, g] ∧ R[rs0, f, g]

Rule argument

Additional information

Used simplifier rules: ⊦ H[r0, o][r] = (r = r0 ∧ r0 ∈ H ⊃ o;H[r]); (s)% ⊦ r ∈ rs → mk(rs, f)[r] = f(r); (s)%r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)% ⊦ ¬ r ∈ ∅; (s)% ⊦ r ∈ H[r0, o] ↔ r ∈ H; (s)% ⊦ r ∈ mk(rs, f) ↔ r ∈ rs; (s)