[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 18 from proof from-succ (previous step)

⊦ nf(n0 + n) = (n0 = 0 ⊃ nf(n);nf(n + 1 + (n0 - 1)))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 19:
n0 ≠ 0 ⊦ nf(n0 + n) = nf(n + 1 + (n0 - 1))

Rule argument

pos: right 1

Additional information

Used simplifier rules: add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2