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Proof step 24 from proof valid-wl-erasing-invalidate0 (previous step)

# apebs = # wlar, apebs[m] = mkpeb(aehdr0, avhdr1, buf1, m1, boolvar), wlar[n] = wl-entry(m5, used), apebs[n] = mkpeb(aechdr(m5), avhdr0, buf0, m3, false), valid-sqns(v, m0, apebs) = bag ++ avhdr1.sqn ++ avhdr0.sqn, n < # wlar, m < # wlar, valid-peb(v, m0, mkpeb(aehdr0, avhdr1, buf1, m1, boolvar)), valid-peb(v, m0, mkpeb(aechdr(m5), avhdr0, buf0, m3, false)), (bag ++ avhdr1.sqn).max ≤ avhdr0.sqn ⊦ avhdr0.sqn = (avhdr0.sqn = avhdr1.sqn ⊃ bag ++ avhdr1.sqn;bag ++ avhdr0.sqn).max, n = m, valid-peb(mkpeb(aehdr, avhdr, buf, m2, false))

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m = (bag ++ m).max ↔ ¬(bag ≠ ∅ ∧ m < bag.max); (s)% ⊦ bag ≠ ∅ → (bag ++ m).max = (bag.max > m ⊃ bag.max;m); (s)% ⊦ m > n ↔ n < m; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ ¬ n < n; (s)