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Proof step 3 from proof eq-01 (previous step)

∀ n, n0. m0 + n = m0 + n0 → n = n0, m0 +1 + n = m0 +1 + n0 ⊦ n = n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m0 + n = m0 + n0, ∀ n, n0. m0 + n = m0 + n0 → n = n0 ⊦ n = n0

Rule argument

Additional information

Used simplifier rules: rec: ⊦ m +1 + n = (m + n)+1;%inj-1: ⊦ n +1 = n0 +1 ↔ n = n0; from specification nat-basic1