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Proof step 154 from proof recovery (previous step)

# apebs0 = # wlar0, # wlar = # wlar0, apebs0[unmapped.pnum] = mkpeb(aehdr, avhdr, buf, m0, boolvar), vols0[v][m] = unmapped, (v × m ∈ q0 ⊃ unmapped;erased).mapped?, m < # vols0[v], abs(avols0, apebs0, vols, q), bflipset-cons(bflipset, wlar), bflipset-cons(bflipset0, wlar0), ebm-cons(avols0), is-max(apebs0, vols0, wlar0, q0, bflipset0, vtblpnum3, sqnum), ubi-io-inv(apebs0), unique-sqns(apebs0), valid-ebatbl(apebs0, wlar, vols, vtblpnum), valid-ebatbl(apebs0, wlar0, vols0, vtblpnum3), valid-eraseq(q), valid-eraseq(wlar, q), valid-eraseq(apebs0, q), valid-eraseq(q0), valid-eraseq(wlar0, q0), valid-eraseq(apebs0, q0), valid-global-sqn(apebs0, vtblpnum0), valid-global-sqn(apebs0, sqnum), valid-vtbl(apebs0, vols, wlar, vtblpnum), valid-vtbl(apebs0, vols0, wlar0, vtblpnum3), valid-wl(apebs0, wlar), valid-wl(apebs0, wlar0), v ∈ vols0, q0.lebrefs ⊆ q.lebrefs, vols ⊆ vols0, vols0⌈vols, q⌉ ⊦ VTBL_VOLID ∈ vols0, VTBL_VOLID ∈ vols

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 5

Additional information

Used simplifier rules: ⊦ ¬ erased.mapped?; (s)% ⊦ ¬ unmapped.mapped?; (s)