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Proof step 77 from proof lift-unknown (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 78:
rs1 ∩ rs0 = ∅, R[rs0, f, g], r ∈ rs1, ∀ r. r ∈ rs1 → g(r) = a, ∀ r0. (r0 ∈ rs1 → r0 = r) ∧ (r0 ∈ rs1 → f(r0) = (r0 = r ⊃ f(r);(∅ ++ r)[r0])) ⊦ # rs1 = 0, g(r) = a, r = null, rs1 = ∅, ∃ o, a. (r ==>(o, a) * R)[rs1 ∪ rs0, f, g]

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]%ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset]