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Proof step 15 from proof firstn-last-01 (previous step)

# x0 = n1 + # z, a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, a0 ∈ firstn(n1, a0 ' + x0), # z ≤ # x0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(# x0 - # z, x0).last = a ⊦ firstn(n1, x0).last = a, n1 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 16:
# x0 = n1 + # z, a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, a0 ∈ firstn(n1, a0 ' + x0), # z ≤ # x0 ⊦ firstn(n1, x0).last = a, n1 = 0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z
Proof step 17:
# x0 = n1 + # z, a0 ' + x0 = firstn(n1, a0 ' + x0) + a ' + z, firstn(# x0 - # z, x0).last = a, a0 ∈ firstn(n1, a0 ' + x0), # z ≤ # x0 ⊦ firstn(n1, x0).last = a, n1 = 0

Rule argument

pos: left 5

Proof step 15 from proof firstn-last-01 (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information