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Proof step 75 from proof lift-unknown (previous step)

r0 = r, rs1 ∩ rs0 = ∅, R[rs0, f, g], ∀ 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], (r0 ∈ rs1 ↔ r0 = r) ∧ (r0 ∈ rs1 → f(r0) = (r0 = r ⊃ o;(∅ ++ r)[r0]o) ∧ g(r0) = (r0 = r ⊃ a;(∅ ++ r)[r0]a))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 76:
rs1 ∩ rs0 = ∅, R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a, ∀ r0. (r0 ∈ rs1 ↔ r0 = r) ∧ (r0 ∈ rs1 → f(r0) = (r0 = r ⊃ o;(∅ ++ r)[r0])) ⊦ r = null, ∃ o, a. (r ==>(o, a) * R)[rs1 ∪ rs0, f, g], r ∈ rs1 ∧ f(r) = o ∧ g(r) = a

Proof step 75 from proof lift-unknown (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information