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Proof step 5 from proof snd-eq (previous step)

m - (maxenc(m) * maxenc(m) + maxenc(m)) / 2 = m ⊦ m = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m, (maxenc(m) * maxenc(m) + maxenc(m)) / 2, n] flags:

Additional information

Used simplifier rules: max-nls: ⊦ ¬ m < (maxenc(m) * maxenc(m) + maxenc(m)) / 2;%max-zero: ⊦ maxenc(m) = 0 ↔ m = 0;%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%two-zero-01-01: ⊦ m / 2 = 0 ↔ ¬ ¬(m = 0 ∨ m = 1); from specification nat-div%square-one: ⊦ m * m + m ≠ 1; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%sqzero: ⊦ m * m = 0 ↔ m = 0; from specification nat-mult%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m, (maxenc(m) * maxenc(m) + maxenc(m)) / 2, n