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Proof step 3 from proof same-used-extract (previous step)

# wlar0 = # wlar1, wlar0[n] = wl-entry(m0, used), n < # wlar1, ∀ n. n < # wlar1 → (wlar0[n].status = used ↔ wlar1[n].status = used) ⊦ wlar1[n].status = used

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# wlar0 = # wlar1, wlar0[n] = wl-entry(m0, used), wlar1[n] = wl-entry(m, stat), n < # wlar1, ∀ n. n < # wlar1 → (wlar0[n].status = used ↔ wlar1[n].status = used) ⊦ stat = used

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar1[n], m, stat]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar1[n], m, stat