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Proof step 6 from proof union-at-right-01 (previous step)

rk ∈ st1[rk0, val], ∀ rk, st. rk ∈ st1 → (st ∪ st1)[rk] = st1[rk] ⊦ ((rk = rk0 ⊃ val; (st ∪ st1)[rk])) = ((rk = rk0 ⊃ val; st1[rk]))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
rk ≠ rk0, rk ∈ st1[rk0, val], ∀ rk, st. rk ∈ st1 → (st ∪ st1)[rk] = st1[rk] ⊦ (st ∪ st1)[rk] = ((rk = rk0 ⊃ val; st1[rk]))

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]