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Proof step 104 from proof payload-shift-left (previous step)

(n0 < n ⊃ zbrar[n0];zbrar[n0 + 1]) = mkzentry(key, adr0), zbrar[n2 + 1] = mkzentry(key, adr), 2 + n0 < # zbrar, n + 1 < # zbrar, n0 + 1 < nused, n0 < # zbrar, n0 < nused, n < nused, n0 < n2, n2 < # zbrar, n2 + 1 < # zbrar, n2 < nused, nused < # zbrar, n < # zbrar, n0 + 1 < # zbrar, sorted(zbrar, 0, nused + 1), key ∈ payload(shl(zbrar, n, nused + 1), 0, nused), n ≤ n2 + 1, n ≤ n2, ∀ n1. ¬ n2 < n1 ∧ zbrar[n1].entry? → zbrar[n1].key ≠ key ⊦ payload(shl(zbrar, n, nused + 1), 0, nused)[key] = adr, zbrar[n].entry?

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m0 → m + n < m0 + n0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%m0 < m, m1 < m, m ≤ # zbrar ⊦ sorted(zbrar, 0, m) ∧ zbrar[m0] = mkzentry(key, adr0) → (zbrar[m1] = mkzentry(key, adr1) ↔ m0 = m1 ∧ adr0 = adr1); (ws)% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m + m0 < n → ¬ n + n0 < m; (s)