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Proof step 6 from proof prefixn-to-adrs (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
n ≤ # lax ⊦ firstn(n, lax.to-adr(m)) = (n ≤ # lax ⊃ firstn(n, lax);lax).to-adr(m)

Rule argument

pos: right 1

Additional information

Used simplifier rules: to-adrs-eq: ⊦ lax0.to-adr(n) = lax1.to-adr(n) ↔ lax0 = lax1;