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Proof step 31 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 = ∅, ∃ o. (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], ∃ o. ((r => o))(mk(rs1, f))

User applied rule exists right

And obtained the following new goals

Proof step 32:
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 = ∅, ((r => f(r)))(mk(rs1, f)), ∃ o. ((r => o))(mk(rs1, f)), ∃ o. (g(r)[r => o] * R)[rs1 ∪ rs0, f, g]

Rule argument

pos: right 5 terms: f(r) abort: F discard: F computed: T

Additional information

Used terms: f(r)