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Proof step 2 from proof enc-zero (previous step)

⊦ encode(encode(encode(i), encode(j))) = 0 ↔ i = 0 ∧ j = 0

User applied rule case distinction

And obtained the following new goals

Proof step 3:
encode(encode(encode(i), encode(j))) = 0 ⊦ i = 0 ∧ j = 0
Proof step 4:
i = 0, j = 0 ⊦ encode(encode(encode(i), encode(j))) = 0

Rule argument

pos: right 1

Additional information