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Proof step 255 from proof inv-recover (previous step)

# wlar1 = # wlar, # apebs0 = # wlar, apebs0[vtblpnum1].vidhdr.leb = VTBL_LNUM, wlar1[vtblpnum1].status = used, apebs0[vtblpnum1].vidhdr.vol = VTBL_VOLID, apebs0[m] = mkpeb(aehdr, avhdr, buf0, m0, boolvar), recs0[VTBL_VOLID × VTBL_LNUM] = recovery-entry(m, avhdr.sqn), buf = mkbuf(LEB_SIZE), vtblpnum1 < # apebs0, vtblpnum1 < # wlar1, nx < # wlar, bflipset-cons(bflipset, wlar), bflipset-cons(bflipset1, wlar1), indices-erasing(wlar, nx), is-encoded(to-vtbl(vols1), apebs0[vtblpnum1].data), ubi-io-inv(apebs0), unique-sqns(apebs0), valid-ebatbl(apebs0, wlar1, vols1, vtblpnum1), valid-eraseq(wlar, q), valid-eraseq(apebs0, q1), valid-eraseq(wlar1, q1), valid-eraseq(q1), valid-eraseq-stronger(apebs0, q), valid-global-sqn(apebs0, sqnum), valid-global-sqn(apebs0, vtblpnum5), valid-recs(apebs0, wlar, recs0), valid-recs-sqns(apebs0, recs0), valid-wl(apebs0, wlar), valid-wl(apebs0, wlar1), VTBL_VOLID × VTBL_LNUM ∈ recs0 ⊦ m = vtblpnum1, dups(nx), VTBL_VOLID ∈ vols1, vtblpnum1 ∈ q1, m ∈ q

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic