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Proof step 5 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, m + 1, nused0 +1)[m, load(brar[m])] = load(brar, m, nused0 +1)[m + nused0 +1, load(brar[m + nused0 +1])]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ brar, m. load(brar, m + 1, nused0)[m, load(brar[m])] = load(brar, m, nused0)[m + nused0, load(brar[m + nused0])] ⊦ load(brar, m + 1, nused0 + 1)[m, load(brar[m])] = load(brar, m, nused0 + 1)[m + nused0 + 1, load(brar[m + nused0 + 1])]

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)