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Proof step 17 from proof get-known (previous step)

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])) ⊦ g(r) = a, r = null

User applied rule all left

And obtained the following new goals

Proof step 18:
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 ⊦ g(r) = a, r = null

Rule argument

pos: left 4 terms: r abort: F discard: F computed: T

Additional information

Used terms: r