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Proof step 37 from proof lift-unknown (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], r ∈ rs1, ∀ r1. (r1 ∈ rs1 → r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ f(r);(∅ ++ r)[r1]o) ∧ g(r1) = (r1 = r ⊃ g(r);(∅ ++ r)[r1]a)) ⊦ # rs1 = 0, r = null, rs1 = ∅, ∀ r0. (r0 ∈ mk(rs1, f) ↔ r0 ∈ (∅ ++ r)[r, f(r)]) ∧ (r0 ∈ mk(rs1, f) → mk(rs1, f)[r0] = (∅ ++ r)[r, f(r)][r0]), ∃ o. (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], ∃ o. ((r => o))(mk(rs1, f))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
rs1 ∩ rs0 = ∅, R[rs0, f, g], r ∈ rs1, ∀ r1. (r1 ∈ rs1 → r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ f(r);(∅ ++ r)[r1]o) ∧ g(r1) = (r1 = r ⊃ g(r);(∅ ++ r)[r1]a)) ⊦ # rs1 = 0, r = null, rs1 = ∅, ∃ o. ((r => o))(mk(rs1, f)), ∃ o. (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], (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)