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Proof step 7 from proof keys-drop-lower (previous step)

m + n + nused0 < # zbrar, m + n < # zbrar[m, zbr], ∀ m, n, zbrar, zbr. m + n + nused0 ≤ # zbrar → keys(zbrar[m, zbr], m + n, nused0) = keys(zbrar, m + n, nused0) ∨ n = 0 ⊦ keys(zbrar[m, zbr], (m + n) + 1, nused0) ++ zbrar[m, zbr][m + n].key = keys(zbrar, m + n, nused0 + 1), n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m + n + nused0 < # zbrar, m + n < # zbrar, ∀ m, n, zbrar, zbr. m + n + nused0 ≤ # zbrar → keys(zbrar[m, zbr], m + n, nused0) = keys(zbrar, m + n, nused0) ∨ n = 0 ⊦ keys(zbrar[m, zbr], (m + n) + 1, nused0) ++ zbrar[m + n].key = keys(zbrar, (m + n) + 1, nused0) ++ zbrar[m + n].key, n = 0

Rule argument

Additional information

Used simplifier rules: keys-rec: ⊦ m < # zbrar → keys(zbrar, m, nused + 1) = keys(zbrar, m + 1, nused) ++ zbrar[m].key;% ⊦ # zbrar[n, zbr] = # zbrar; (s)%m ≠ n ⊦ zbrar[m, zbr][n] = zbrar[n]; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s)