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Proof step 11 from proof valid-eraseqp-valid-lebref (previous step)

apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # apebs, valid-eraseq(apebs, q0), valid-peb(v, m, apebs[n]), valid-peb(mkpeb(aehdr, avhdr, buf, m1, boolvar)) → lref = avhdr.vol × avhdr.leb, ∀ apebs, v, m, n. valid-eraseq(apebs, q0) ∧ valid-peb(v, m, apebs[n]) ∧ n ∈ q0 → eq-entry(n, v × m) ∈ q0, n ∈ eq-entry(m0, lref) ' ∨ n ∈ q0 ⊦ eq-entry(n, v × m) = eq-entry(m0, lref), eq-entry(n, v × m) ∈ q0

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: apebs, v, m, n abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ n ∈ eq-entry(m, lref) ' ↔ n = m; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)% ⊦ eq-entry(m, lref) = eq-entry(m0, lref0) ↔ m = m0 ∧ lref = lref0; (ws)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ valid-peb(v, m, apeb) → valid-peb(apeb); (ss)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.vol = v; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.leb = m; (s) Used terms: apebs, v, m, n