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Proof step 3 from proof inj (previous step)

∀ m. n→i(m) = n→i(n0) → m = n0, n→i(m) = n→i(n0 +1) ⊦ m = n0 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n→i(m) = n→i(n0) + 1, ∀ m. n→i(m) = n→i(n0) → m = n0 ⊦ m = n0 + 1

Rule argument

Additional information

Used simplifier rules: nat2int-03: ⊦ n→i(n + 1) = n→i(n) + 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2