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Proof step 3 from proof eq-natinf-to-nat-2-1 (previous step)

⊦ ⌜n + m⌝ = ⌜n0⌝ ↔ n + m = n0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-inf-basic inst: name: inj seq: ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0 subst: [m0, m] → [n0, n + m] flags: rewrite

Additional information

Used terms: n0, n + m