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Proof step 10 from proof program+leftmost (previous step)

((r0 = r ⊃ mknode(r4, a0, r1); mknode(r2, a2, r4))) = mknode(r3, a1, r5), H[r] = mknode(r0, a0, r1), H[r0] = mknode(r2, a2, r4), r0 ∈ H, r ∈ H, (r0 => mknode(null, a, r6) * tr(r1, t1) * tr(r6, t2) * r => mknode(r4, a0, r1) * P)(H[r, mknode(r4, a0, r1)]), ¬ (tr(r, branch(t2, a0, t1)) * P * true)(H[r, mknode(r4, a0, r1)]) ∨ a2 ≠ a ⊦ r4 = r5 ∧ a2 = a1

User applied rule case distinction

And closed the goal!

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ mknode(r, a, r0) = mknode(r1, a0, r2) ↔ r = r1 ∧ a = a0 ∧ r0 = r2], [CSIMPSEQ : ⊦ (P0 * P1) * P2 = P0 * P1 * P2], [CSIMPSEQ : ⊦ P0 * P1 = P1 * P0])]]