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Proof step 1 from proof nat2int-05 (conclusion)

⊦ n→i(m) = n→i(n) + 1 ↔ m = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n→i(m) = n→i(n) + 1 ⊦ m = n + 1

Rule argument

Additional information

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