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Proof step 7 from proof in-playload-data (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
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)) ⊦ adr ∈ payload(zbrar, n, m0 + 1) ↔ (∃ m. n ≤ m ∧ ¬ n + m0 < m ∧ zbrar[m].entry? ∧ zbrar[m].adr = adr)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)