[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 26 from proof union-def (previous step)

∀ rk, st, val. rk ∈ st ∪ st1 ∧ (st ∪ st1)[rk] = val ↔ ¬ rk ∈ st1 ∧ rk ∈ st ∧ st[rk] = val ∨ rk ∈ st1 ∧ st1[rk] = val ⊦ rk ∈ (st ∪ st1)[rk0, val0] ∧ ((rk = rk0 ⊃ val0; (st ∪ st1)[rk])) = val ↔ ¬ rk ∈ st1[rk0, val0] ∧ rk ∈ st ∧ st[rk] = val ∨ (rk = rk0 ∨ rk ∈ st1) ∧ ((rk = rk0 ⊃ val0; st1[rk])) = val

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 18:
rk ∈ (st ∪ st1)[rk0, val0] ∧ ((rk = rk0 ⊃ val0; (st ∪ st1)[rk])) = val, ∀ rk, st, val. rk ∈ st ∪ st1 ∧ (st ∪ st1)[rk] = val ↔ ¬ rk ∈ st1 ∧ rk ∈ st ∧ st[rk] = val ∨ rk ∈ st1 ∧ st1[rk] = val ⊦ ¬ rk ∈ st1[rk0, val0] ∧ rk ∈ st ∧ st[rk] = val ∨ (rk = rk0 ∨ rk ∈ st1) ∧ ((rk = rk0 ⊃ val0; st1[rk])) = val
Proof step 25:
¬ rk ∈ st1[rk0, val0] ∧ rk ∈ st ∧ st[rk] = val ∨ (rk = rk0 ∨ rk ∈ st1) ∧ ((rk = rk0 ⊃ val0; st1[rk])) = val, ∀ rk, st, val. rk ∈ st ∪ st1 ∧ (st ∪ st1)[rk] = val ↔ ¬ rk ∈ st1 ∧ rk ∈ st ∧ st[rk] = val ∨ rk ∈ st1 ∧ st1[rk] = val ⊦ rk ∈ (st ∪ st1)[rk0, val0] ∧ ((rk = rk0 ⊃ val0; (st ∪ st1)[rk])) = val

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]