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Proof step 1 from proof ls-01 (conclusion)

n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 2:
n ≤ m ⊦ n0 = m - n ↔ m = n0 + n

Rule argument

spec: <toplevel> name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, n0, n1] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n, n0, n1