[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 7 from proof addex (previous step)

m < n0 → m + m0 = n0 ∧ m0 ≠ 0, ∀ m1. m1 < n0 → (∃ m0. m1 + m0 = n0 ∧ m0 ≠ 0) ⊦ n0 = 0, n0 < m, ∃ m0. m + m0 = n0 + 1 ∧ m0 ≠ 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: n0 ≠ m

Additional information

Used simplifier rules: eq: ⊦ m = m + n ↔ n = 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%l-01: ⊦ m < m + n ↔ n ≠ 0;%eq-01: ⊦ m + n = m + n0 ↔ n = n0;%oz: ⊦ m + 1 ≠ 0;%eq-01: ⊦ m + n = m + n0 ↔ n = n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%ref-1: ⊦ ¬ n < n; from specification nat-basic1%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1 Used formulas: n0 ≠ m