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Proof step 5 from proof le-add-01 (previous step)

n < m, m + m1 ≤ n ⊦ m = 0, n0 + m1 = 0, m1 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)

Rule argument

spec: nat-basic2 inst: name: addex seq: ⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0) subst: [n, m] → [m, n] flags: rewrite

Additional information

Used simplifier rules: le-add: ⊦ m + m0 ≤ m ↔ m0 = 0;%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m, n