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Proof step 8 from proof lastN-ex (previous step)

# x = n0 + n, x = firstn(n0, x) + restn(n0, x) ⊦ n = 0, ∃ y. x = y + lastn(n, x) ∧ # y = n0

User applied rule exists right

And obtained the following new goals

Proof step 9:
# x = n0 + n, x = firstn(n0, x) + restn(n0, x) ⊦ n = 0, x = firstn(n0, x) + lastn(n, x) ∧ # firstn(n0, x) = n0, ∃ y. x = y + lastn(n, x) ∧ # y = n0

Rule argument

pos: right 2 terms: firstn(n0, x) abort: F discard: F computed: F

Additional information

Used terms: firstn(n0, x)