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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;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2