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Proof step 4 from proof prefixn-to-adrs (previous step)

⊦ (n ≤ # lax ⊃ firstn(n, lax.to-adr(m));lax.to-adr(m)) = prefixn(n, lax).to-adr(m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ prefixn(n, lax) = (n ≤ # lax ⊃ firstn(n, lax);lax)
Proof step 6:
⊦ (n ≤ # lax ⊃ firstn(n, lax.to-adr(m));lax.to-adr(m)) = (n ≤ # lax ⊃ firstn(n, lax);lax).to-adr(m)

Rule argument

spec: list-prefixn, inst: laddress-list name: prefixn seq: ⊦ prefixn(n, lax) = (n ≤ # lax ⊃ firstn(n, lax);lax) subst: [lax, n] → [lax, n] flags: <all paths>

Additional information

Used terms: lax, n