[Imprint][Data Privacy Statement][E-Mail]



View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 9 from proof le-sum (previous step)

m0 < n1, n0 + m0 < n + n1, n ≤ n0, n1 ≤ m0 + 1, m0 + n0 + 1 < n + n1, ∀ n, n0, n1. n ≤ n0 ∧ n1 ≤ m0 → n + n1 ≤ n0 + m0 ⊦ n1 = 0, n + n1 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat inst: name: anti seq: ⊦ m ≤ n → (n ≤ m ↔ m = n) subst: [m, n] → [m0 + 1, n1] flags: rewrite, current

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m;%succ-01: ⊦ m + 1 ≤ n ↔ m < n;%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false);%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false);%oz: ⊦ m + 1 ≠ 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%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%pls: m < m0, n < n0 ⊦ m + n + 1 < m0 + n0; from specification nat-basic2 Used terms: m0 + 1, n1