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Proof step 29 from proof aux-known (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
rs1 ∩ rs0 = ∅, R[rs0, f, g], r ∈ rs0, ∀ r. r ∈ rs1 → g(r) = a0, ∀ r1. (r1 ∈ rs1 ↔ r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ o;(∅ ++ r)[r1])) ⊦ # rs0 = 0, r = null, r0 = r, rs0 = ∅, R[rs0, f, g(r; a1)], r0 ∈ rs1, (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]% ⊦ rs ∩ rs0 = ∅ ∧ r ∈ rs → (r ∈ rs0 ↔ false); (as)%int-c: ⊦ rs ∩ rs0 = rs0 ∩ rs; from specification set-union[refset]%int-a: ⊦ (rs ∩ rs0) ∩ rs1 = rs ∩ rs0 ∩ rs1; from specification set-union[refset]%ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset]