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Proof step 25 from proof aux-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], r0 ∈ rs1 → g(r0) = a0, ∀ r. r ∈ rs1 → g(r) = a0, ∀ r1. (r1 ∈ rs1 ↔ r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ o;(∅ ++ r)[r1])) ⊦ r = null, (r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1) ∧ R[rs0, f, g(r; a1)], (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0

User applied rule all left

And obtained the following new goals

Proof step 26:
r0 ∈ rs1 ↔ r0 = r, rs1 ∩ rs0 = ∅, R[rs0, f, g], r0 ∈ rs1 → g(r0) = a0, r0 ∈ rs1 → f(r0) = (r0 = r ⊃ o;(∅ ++ r)[r0]), ∀ r1. (r1 ∈ rs1 ↔ r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ o;(∅ ++ r)[r1])), ∀ r. r ∈ rs1 → g(r) = a0 ⊦ r = null, (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0, (r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1) ∧ R[rs0, f, g(r; a1)]

Rule argument

pos: left 5 terms: r0 abort: F discard: F computed: T

Additional information

Used terms: r0