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Proof step 2 from proof snd-ls (previous step)

m ≤ m.2 ⊦ m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: dec-snd seq: ⊦ m.2 = m - (maxenc(m) * (maxenc(m) + 1)) / 2 subst: [m] → [m] flags: <all paths>

Additional information

Used simplifier rules: max-nls: ⊦ ¬ m < (maxenc(m) * maxenc(m) + maxenc(m)) / 2;%max-zero: ⊦ maxenc(m) = 0 ↔ m = 0;% ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m ≤ n ↔ ¬ n < m; (ss)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ m / 2 = 0 ↔ m = 0 ∨ m = 1; (s)% ⊦ m * m + m ≠ 1; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)% ⊦ m * m = 0 ↔ m = 0; (s) Used terms: m