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Proof step 5 from proof union-def (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 4:
rk ≠ rk0, rk ∈ (st ∪ st1)[rk0, val0], rk ∈ st1[rk0, val0], ∀ rk, st, val. rk ∈ st ∪ st1 ∧ (st ∪ st1)[rk] = val ↔ ¬ rk ∈ st1 ∧ rk ∈ st ∧ st[rk] = val ∨ rk ∈ st1 ∧ st1[rk] = val ⊦ st1[rk] = ((rk = rk0 ⊃ val0; (st ∪ st1)[rk]))

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]