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Proof step 36 from proof in-playload-data (previous step)

n < # zbrar, n + m0 < # zbrar, sorted(zbrar, 0, n + m0 + 1), ∀ zbrar, n, adr. sorted(zbrar, 0, n + m0) ∧ n + m0 ≤ # zbrar → (adr ∈ payload(zbrar, n, m0) ↔ (∃ m1. n ≤ m1 ∧ m1 < n + m0 ∧ zbrar[m1].entry? ∧ zbrar[m1].adr = adr)), ∃ m. n ≤ m ∧ ¬ n + m0 < m ∧ zbrar[m].entry? ∧ zbrar[m].adr = adr ⊦ adr ∈ (zbrar[n].entry? ⊃ payload(zbrar, n + 1, m0)[zbrar[n].key, zbrar[n].adr];payload(zbrar, n + 1, m0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
zbrar[m].entry?, m < # zbrar, n + m0 < # zbrar, n < # zbrar, sorted(zbrar, 0, n + m0 + 1), m ≤ n + m0, n ≤ m, ∀ zbrar, n, adr. sorted(zbrar, 0, n + m0) ∧ n + m0 ≤ # zbrar → (adr ∈ payload(zbrar, n, m0) ↔ (∃ m1. n ≤ m1 ∧ m1 < n + m0 ∧ zbrar[m1].entry? ∧ zbrar[m1].adr = adr)) ⊦ zbrar[m].adr ∈ (zbrar[n].entry? ⊃ payload(zbrar, n + 1, m0)[zbrar[n].key, zbrar[n].adr];payload(zbrar, n + 1, m0))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)