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Proof step 13 from proof load-branches-last (previous step)

∀ brar, m. load(brar, m + 1, nused0)[m, load(brar[m])] = load(brar, m, nused0)[m + nused0, load(brar[m + nused0])] ⊦ load(brar, 2 + m, nused0)[m + 1, load(brar[m + 1])][m, load(brar[m])] = load(brar, m + 1, nused0)[m + nused0 + 1, load(brar[m + nused0 + 1])][m, load(brar[m])], m = m + nused0 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
∀ brar, m. load(brar, m + 1, nused0)[m, load(brar[m])] = load(brar, m, nused0)[m + nused0, load(brar[m + nused0])] ⊦ load(brar, 2 + m, nused0)[m + 1, load(brar[m + 1])][m, load(brar[m])] = load(brar, m + 1, nused0)[m + nused0 + 1, load(brar[m + nused0 + 1])][m, load(brar[m])]

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 + 1 ≠ 0; (s)% ⊦ m = m + n ↔ n = 0; (s)