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Proof step 15 from proof lst-in-hp (previous step)

x = a + x0, rs1 ∩ rs0 = ∅, R[rs0, cf, af], ls(r, x)(mk(rs1, cf)), ∀ r. r ∈ rs1 → af(r) = o ⊦ r = null, r ∈ rs1, r ∈ rs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
rs1 ∩ rs0 = ∅, ((r => a × r0 * ls(r0, x0)))(mk(rs1, cf)), R[rs0, cf, af], ∀ r. r ∈ rs1 → af(r) = o ⊦ r = null, r ∈ rs0, r ∈ rs1

Rule argument

Additional information

Used simplifier rules: ls-rec: ⊦ ls(r, a ' + x) = _∃(λ r0. r => a × r0 * ls(r0, x));% ⊦ (_∃ rP)(H) ↔ (∃ r. rP(r)(H)); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)