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Proof step 5 from proof del-lists-1 (previous step)

⊦ del-list(r + [], n) = _∃(λ x. ls(r, x) ∧ ⌜n = # x⌝ ), r = null

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ r + rx = r ' + rx
Proof step 7:
⊦ del-list(r ' + [], n) = _∃(λ x. ls(r, x) ∧ ⌜n = # x⌝ ), r = null

Rule argument

spec: list, inst: reflist name: c seq: ⊦ r + rx = r ' + rx subst: [rx, r] → [[], r] flags: <all paths>

Additional information

Used terms: [], r