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Proof step 2 from proof firstn-prefix-03 (previous step)

# x = (1 + # z) + n0, a ' + z ⊒ x ⊦ firstn(n0, x) ⊑ x ↔ true

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# x = (1 + # z) + n0, a ' + z ⊒ x ⊦ firstn(n0, x) ⊑ x

Rule argument

Additional information