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Proof step 5 from proof sw-put (previous step)

r1 ∈ H, r0 ∈ H ⊦ r0 = r1, ∀ r. (r ∈ H[r0, o0][r1, o1] ↔ r ∈ H[r1, o1][r0, o0]) ∧ (r ∈ H[r0, o0][r1, o1] → H[r0, o0][r1, o1][r] = H[r1, o1][r0, o0][r])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
r0 ∈ H, r1 ∈ H, r ∈ H ⊦ (r = r1 ⊃ o1;(r = r0 ⊃ o0;H[r])) = (r = r0 ⊃ o0;(r = r1 ⊃ o1;H[r])), r0 = r1

Rule argument

Additional information

Used simplifier rules: in-write: ⊦ r ∈ H[r0, o] ↔ r ∈ H;%at-write: ⊦ H[r0, o][r] = (r = r0 ∧ r0 ∈ H ⊃ o;H[r]);