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Proof step 12 from proof same-used2-idexpt (previous step)

# wlar1 = # wlar2, # wlar0 = # wlar2, stat1 = used ↔ wlar1[n].status = used, wlar2[n] = wl-entry(m3, stat2), wlar1[m] = wl-entry(m0, stat), wlar2[m] = wl-entry(m1, stat0), wlar0[n] = wl-entry(m2, stat1), n < # wlar2, n ≠ m → wlar1[n] = wl-entry(m3, stat2) ⊦ stat1 = used ↔ stat2 = used, stat = used, stat0 = used

Heuristic elimination applied rule insert elim lemma

And closed the goal!

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], m4, stat3]

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat) = wl-entry(m0, stat0) ↔ m = m0 ∧ stat = stat0; (ws)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar1[n], m4, stat3