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Proof step 2 from proof nat2int-05 (previous step)

n→i(m) = n→i(n) + 1 ⊦ m = n + 1

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: m

Additional information

Used simplifier rules: nat2int-07: ⊦ 0 ≠ n→i(m) + 1;%succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%inj: ⊦ n→i(m) = n→i(n) ↔ m = n;%nat freely generated by 0, +1;;% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ i + j = i + i0 ↔ j = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ m + n = m + n0 ↔ n = n0; (s)