[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 24 from proof refs-equal (previous step)

¬ r0 ≠ rk .ref, # st .refs = # st0 .refs, r0 ∈ st0 .refs, r0 ∈ st .refs, r ∈ st0, rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], rk ∈ st -- r0, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r = r0, st0 = ∅, st = ∅, r ∈ st, rk ∈ st0 -- r0 ∧ (st -- r0)[rk] = (st0 -- r0)[rk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
# st .refs = # st0 .refs, rk .ref ∈ st0 .refs, rk .ref ∈ st .refs, r ∈ st0, rk ∈ st -- rk .ref, rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r = rk .ref, st0 = ∅, st = ∅, r ∈ st, rk ∈ st0 -- rk .ref ∧ (st -- rk .ref)[rk] = (st0 -- rk .ref)[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]