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Proof step 28 from proof lift-unknown (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 29:
r ∈ rs1 ↔ r = r, rs1 ∩ rs0 = ∅, R[rs0, f, g], r ∈ rs1 → f(r) = (r = r ⊃ o;(∅ ++ r)[r]o) ∧ g(r) = (r = r ⊃ a;(∅ ++ r)[r]a), ∀ r1. (r1 ∈ rs1 ↔ r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ o;(∅ ++ r)[r1]o) ∧ g(r1) = (r1 = r ⊃ a;(∅ ++ r)[r1]a)) ⊦ r = null, ∃ o. (a[r => o] * R)[rs1 ∪ rs0, f, g], (∃ o. ((r => o))(mk(rs1, f))) ∧ (r0 ∈ rs1 → g(r0) = a)

Rule argument

pos: left 3 terms: r abort: F discard: F computed: T

Additional information

Used terms: r