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Proof step 4 from proof payload-drop-upper (previous step)

m + nused0 +1 ≤ # zbrar, n ≤ # zbrar, m + nused0 +1 ≤ n, ∀ m, n, zbrar, zbr. m + nused0 ≤ n ∧ n ≤ # zbrar ∧ m + nused0 ≤ # zbrar → payload(zbrar[n, zbr], m, nused0) = payload(zbrar, m, nused0) ⊦ payload(zbrar[n, zbr], m, nused0 +1) = payload(zbrar, m, nused0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m + nused0 < # zbrar, m + nused0 < n, n ≤ # zbrar, ∀ m, n, zbrar, zbr. m + nused0 ≤ n ∧ n ≤ # zbrar ∧ m + nused0 ≤ # zbrar → payload(zbrar[n, zbr], m, nused0) = payload(zbrar, m, nused0) ⊦ payload(zbrar[n, zbr], m, nused0 + 1) = payload(zbrar, m, nused0 + 1)

Rule argument

Additional information

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