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Proof step 29 from proof valid-ebatbl-unmap (previous step)

vols[v1][m0] = ⌜n⌝ , vols[v0][m1] = ⌜n⌝ , apebs[n] = mkpeb(aehdr, avidhdr(v1, m0, m2, m4, m5), buf, m3, boolvar), wlar[n] = wl-entry(m6, used), m1 < # vols[v0], n < # wlar, n < # apebs, m0 < # vols[v1], v1 ∈ vols, v0 ∈ vols, ∀ v, m, n1. vols[v][m] = ⌜n1⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n1].status = used ∧ n1 < # apebs ∧ n1 < # wlar ∧ apebs[n1].vidhdr.valid? ∧ apebs[n1].vidhdr.vol = v ∧ apebs[n1].vidhdr.leb = m ∧ n1 ≠ n0 ⊦ n = n0, v0 = v1

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 11 terms: v0, m1, n

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)% ⊦ avidhdr(v, m, m0, m1, m2).vol = v; (s)% ⊦ avidhdr(v, m, m0, m1, m2).leb = m; (s) Used terms: v0, m1, n