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Proof step 20 from proof switch-array (previous step)

r0 ≠ r, # vals = m + 1, ∀ st, vals0, ty0, ty, r0, r, vals1. # vals1 < m + 1 → (true → r = r0 ∨ addarray(r, ty, vals1, addarray(r0, ty0, vals0, st)) = addarray(r0, ty0, vals0, addarray(r, ty, vals1, st))) ⊦ r = r0, vals = [], (addarray(r0, ty0, vals0, addarray(r, ty, vals .butlast, st)[r -- n→i(m) ', vals .last])[r -- _length, intval(n→i(m) + 1)]) = addarray(r0, ty0, vals0, addarray(r, ty, vals .butlast, st)[r -- n→i(m) ', vals .last][r -- _length, intval(n→i(m) + 1)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# vals = m + 1, ∀ st, vals0, ty0, ty, r0, r, vals1. # vals1 < m + 1 → (true → r = r0 ∨ addarray(r, ty, vals1, addarray(r0, ty0, vals0, st)) = addarray(r0, ty0, vals0, addarray(r, ty, vals1, st))) ⊦ r0 = r, (addarray(r0, ty0, vals0, addarray(r, ty, vals .butlast, st)[r -- n→i(m) ', vals .last])[r -- _length, intval(n→i(m) + 1)]) = addarray(r0, ty0, vals0, addarray(r, ty, vals .butlast, st)[r -- n→i(m) ', vals .last][r -- _length, intval(n→i(m) + 1)]), vals = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]