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Proof step 3 from proof lastN-ex (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ n < # x ↔ x = firstn(n, x) + restn(n, x) ∧ n < # x
Proof step 5:
# x - n < # x ↔ x = firstn(# x - n, x) + restn(# x - n, x) ∧ # x - n < # x, # x = n0 + n ⊦ ∃ y. x = y + lastn(n, x) ∧ # y = n0

Rule argument

spec: list-del inst: name: firstrestex seq: ⊦ n < # x ↔ x = firstn(n, x) + restn(n, x) ∧ n < # x subst: [n, x] → [# x - n, x] flags: current, []

Additional information

Used terms: # x - n, x