[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 4 from proof pe-01 (previous step)

(n0 + m) + n = m2 +1 ∧ (n0 + m) + n ≠ 0, m + n = m1 +1 ∧ m + n ≠ 0, m0 < m ⊦ n0 + m1 = m2, m = 0

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: n

Additional information

Used simplifier rules: z: ⊦ 0 + m = m;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%succ: ⊦ m +1 = m + 1;%eq-01: ⊦ m + n = m + n0 ↔ n = n0;%succ: ⊦ m +1 = m + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%eq-01: ⊦ m + n = m + n0 ↔ n = n0;%c: ⊦ m + n = n + m;%nat freely generated by 0, +1;;%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1