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

# wlar0 = # wlar1, wlar1[m] = wl-entry(m1, stat0), wlar0[m] = wl-entry(m0, stat), n < # wlar1, id○(wlar0, wlar1, m) ⊦ wlar0[n].status = used ↔ wlar1[n].status = used, stat = used, stat0 = used

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# wlar0 = # wlar1, wlar0[m] = wl-entry(m0, stat), wlar1[m] = wl-entry(m1, stat0), wlar0[n] = wl-entry(m2, stat1), n < # wlar1, id○(wlar0, wlar1, m) ⊦ stat1 = used ↔ wlar1[n].status = used, stat = used, stat0 = used

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar0[n], m2, stat1]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar0[n], m2, stat1