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Proof step 118 from proof write-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], ∀ r0. (r0 ∈ rs1 ↔ r0 = r) ∧ (r0 ∈ rs1 → (r = r0 ⊃ o1;f(r0)) = (r0 = r ⊃ o1;(∅ ++ r)[r0])), ∀ r. r ∈ rs1 → g(r) = a ⊦ r = null, (a[r => f(r)] * R)[rs1 ∪ rs0, f, g], r ∈ rs0, ∀ r0. (r0 ∈ mk(rs1, f) ↔ r0 ∈ (∅ ++ r)[r, f(r)]) ∧ (r0 ∈ mk(rs1, f) → mk(rs1, f)[r0] = (∅ ++ r)[r, f(r)][r0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 119:
rs1 ∩ rs0 = ∅, R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a, ∀ r0. (r0 ∈ rs1 ↔ r0 = r) ∧ (r0 ∈ rs1 → (r = r0 ⊃ o1;f(r0)) = (r0 = r ⊃ o1;(∅ ++ r)[r0])) ⊦ r = null, (a[r => f(r)] * R)[rs1 ∪ rs0, f, g], r ∈ rs0, (r0 ∈ rs1 ↔ r0 = r) ∧ (r0 ∈ rs1 → f(r0) = (r0 = r ⊃ f(r);(∅ ++ r)[r0]))

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)